Summary to follow. 9 syllabus statements · 31 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 9 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Position
Position
The location of a body relative to a chosen origin, given by its distance and direction from that origin. For motion along a straight line it is a single signed coordinate, x, measured from the origin in a chosen positive direction. SI unit: m. Together with velocity and acceleration, each known as a function of time, position describes and allows analysis of the motion of a body.
Scalar and vector quantities
A scalar has magnitude only (for example distance, speed and time); a vector has magnitude and direction (for example position, displacement, velocity and acceleration). For motion along a straight line, the direction of a vector is shown by its sign relative to the chosen positive direction, so a negative velocity means motion in the negative direction, not a small or decreasing velocity.
Students often think A positive acceleration means speeding up and a negative acceleration means slowing down. In fact Not necessarily. A body speeds up when its acceleration and velocity have the same direction and slows down when they are opposite; the sign of the acceleration alone only gives its direction.
Students often think A body always moves in the direction of its acceleration. In fact No. A body moves in the direction of its velocity. Its acceleration can point in any direction relative to its velocity, including the opposite direction.
Velocity
Velocity
The rate of change of position, v = Δx/Δt. It is a vector: its magnitude is the speed and its direction is the direction of motion. SI unit: m s⁻¹. On a position–time graph the velocity is the gradient.
Acceleration
The rate of change of velocity, a = Δv/Δt. It is a vector. SI unit: m s⁻². On a velocity–time graph the acceleration is the gradient. A body accelerates whenever its velocity changes in magnitude or direction: if acceleration and velocity have the same direction the speed increases; if they are opposite the speed decreases. A body can have a non-zero acceleration at an instant when its velocity is zero, as at the top of a vertical throw.
Students often think If the velocity of a body is zero, its acceleration must also be zero. In fact Not necessarily. Acceleration is the rate of change of velocity. At the top of a vertical throw the velocity is zero but is changing, so the acceleration is g downwards.
Students often think Acceleration increases when speed increases: a faster body has a greater acceleration, and a body whose speed is still rising has an increasing acceleration. In fact No. Acceleration measures how quickly the velocity changes, not how large the velocity is. A car cruising at a steady 30 m s⁻¹ has zero acceleration.
Displacement
Displacement
The change in position of a body, Δx = x(final) − x(initial). It is a vector, so it has a direction (a sign in one dimension). SI unit: m. It depends only on the start and end positions, not on the route between them. On a velocity–time graph the displacement is the area between the graph and the time axis, with areas below the axis counting as negative.
Students often think The displacement of a body is its final position measured from the origin. In fact No. Position is measured from the origin; displacement is the CHANGE in position, measured from where the body started.
Students often think Displacement is a length, so, like distance, it is always positive. In fact Yes. Displacement is a vector; in one dimension its sign shows its direction relative to the chosen positive direction.
Distance
Distance
The total length of the path travelled by a body. It is a scalar and cannot be negative. SI unit: m. Distance depends on the route taken and can never be less than the magnitude of the displacement; the two are equal only for motion in one direction along a straight line.
Students often think Displacement is the same as the total distance travelled along the path. In fact Only for motion in one direction along a straight line. In general the distance is the path length and the magnitude of the displacement is the straight-line change in position, which can be smaller, even zero.
Students often think Distance is the straight-line separation between the start and end points, and displacement is the length of the path followed. In fact No. In physics, distance is the length of the path actually travelled. The straight-line change in position from start to end is the displacement.
Speed
Speed
The rate of change of distance travelled. It is a scalar. SI unit: m s⁻¹. The instantaneous speed is the magnitude of the instantaneous velocity.
Average velocity
The displacement divided by the time interval over which it occurs, Δx/Δt. SI unit: m s⁻¹. On a position–time graph it is the gradient of the straight line (chord) joining the start and end points of the interval. A body that returns to its starting point has zero average velocity.
Average speed
The total distance travelled divided by the total time taken. SI unit: m s⁻¹. It is generally not equal to the magnitude of the average velocity, and it is not the arithmetic mean of the speeds unless equal times are spent at each speed.
Instantaneous velocity
The velocity at a particular instant: the limit of Δx/Δt as the time interval tends to zero. On a position–time graph it is the gradient of the tangent to the graph at that instant. SI unit: m s⁻¹. Its magnitude is the instantaneous speed.
Average and instantaneous acceleration
The average acceleration over an interval is the change in velocity divided by the time taken, Δv/Δt, the gradient of the chord on a velocity–time graph. The instantaneous acceleration is the gradient of the tangent to the velocity–time graph at that instant. SI unit: m s⁻². For uniform acceleration the two are equal at every instant.
Students often think The average speed of a journey is the arithmetic mean of the speeds on its separate parts. In fact Only if equal TIMES are spent at each speed. Average speed is total distance divided by total time.
Students often think The velocity or acceleration at an instant is the value read off the graph at that instant, whatever quantity is plotted. In fact No. The gradient is the rate at which the plotted quantity changes. The value (height) of the graph at a point is the plotted quantity itself.
Equations of motion for uniform acceleration
Equations of motion for uniform acceleration
For a body with constant acceleration a, initial velocity u, final velocity v, displacement s and time t: s = ((u + v)/2)t; v = u + at; s = ut + ½at²; v² = u² + 2as. They are valid only while the acceleration is constant in magnitude and direction, and every vector quantity (s, u, v, a) must be given a sign using one consistent positive direction. They remain valid through an instant of zero velocity, such as the top of a vertical throw.
Students often think The displacement due to the acceleration is at², so s = ut + at² (for a drop from rest, s = gt²). In fact No. The ½ is part of the equation: it arises because, under uniform acceleration from u, the average velocity over time t is u + ½at.
Students often think The term ½at² means ½ × a × t, or the square is applied carelessly, so t is not squared. In fact The time. The term is ½ × a × t², with the time squared and the acceleration not.
Uniform acceleration
Uniform acceleration
Acceleration that is constant in magnitude and direction. The velocity–time graph is a straight line, and the velocity changes by equal amounts in equal time intervals. Free fall near the Earth's surface with negligible fluid resistance is the standard example.
Non-uniform acceleration
Acceleration that changes with time. The velocity–time graph is curved; the instantaneous acceleration is the gradient of the tangent and the displacement is still the area under the graph. The equations of motion cannot be applied, and the average velocity is not in general (u + v)/2. A body falling through air, whose acceleration decreases as fluid resistance grows, is an example.
Students often think The average velocity during any motion is the mean of the initial and final velocities, (u + v)/2. In fact Only when the acceleration is uniform. For non-uniform acceleration the average velocity is the displacement divided by the time, which can be greater or less than (u + v)/2.
Students often think Distance = speed × time can be applied to an accelerating body using its initial or final speed. In fact Only if the average speed is used. Using the initial or final speed treats the body as though it moved at that speed throughout.
Acceleration of free fall, g
Acceleration of free fall, g
The acceleration of a body moving under gravity alone, with no fluid resistance. Near the Earth's surface it is about 9.8 m s⁻², directed vertically downwards, and it is the same for all bodies whatever their mass. In this course g is taken as constant for projectile problems.
Projectile
A body that, after launch, moves under the action of gravity alone (fluid resistance absent or negligible). Its horizontal velocity component is constant and its vertical motion has constant acceleration g downwards. Its trajectory is a parabola; the equation of that parabola is not required.
Components of the launch velocity
A launch velocity u at angle θ to the horizontal is resolved into a horizontal component u cos θ and a vertical component u sin θ. For a launch above the horizontal the vertical component is upwards; for a launch below the horizontal it is downwards; for a horizontal launch it is zero. The equations of motion are then applied separately to each component.
Independence of horizontal and vertical motion
In the absence of fluid resistance, the horizontal and vertical motions of a projectile are independent: the horizontal velocity stays constant (zero horizontal acceleration) while the vertical velocity changes at g. The time of flight is set by the vertical motion alone, so a body launched horizontally and one dropped from rest at the same height reach the ground at the same time.
Time of flight, maximum height and range
The time of flight is the time from launch to landing, found from the vertical motion. At the maximum height the vertical component of velocity is zero but the horizontal component is not. The range is the horizontal distance travelled, equal to the (constant) horizontal velocity component multiplied by the time of flight.
Students often think Heavier bodies fall faster than lighter ones, even when fluid resistance is negligible. In fact No. In the absence of fluid resistance all bodies fall with the same acceleration, g, whatever their mass.
Students often think A body moving horizontally stays up longer than one dropped from rest, because its horizontal motion holds it up. In fact No. Without fluid resistance the horizontal and vertical motions are independent; the time to fall is set by the vertical motion alone.
Fluid resistance
Fluid resistance
The resistive force exerted by a gas or a liquid on a body moving through it. It acts opposite to the velocity of the body relative to the fluid and increases as the speed increases, so it is not constant during a motion.
Terminal speed
The constant speed eventually reached by a body falling through a fluid, when the upward fluid resistance (together with any buoyancy force) has grown to equal the weight. The resultant force and the acceleration are then zero, so the speed stays constant.
Effect of fluid resistance on a projectile
Compared with the same launch without fluid resistance: the horizontal velocity component decreases during the flight; the acceleration is not constant and is not g; the maximum height and the range are reduced; for a launch and landing at the same level the time of flight is reduced; the trajectory is no longer a symmetrical parabola, because the descent is steeper than the ascent and takes longer. A body that falls far enough approaches terminal speed and a path that is close to vertical.
Students often think Fluid resistance makes a projectile's path smaller, but the path is still a symmetrical parabola. In fact No. Fluid resistance reduces the horizontal velocity throughout the flight and the body descends more steeply than it rose, so the highest point is more than halfway along the range and the descent takes longer than the ascent.
Students often think Fluid resistance (air resistance) is a fixed force for a given body, independent of its speed. In fact No. Fluid resistance increases with the speed of the body relative to the fluid, so it changes during the motion.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 A body moves along a straight line, with positive taken to the right. At one instant its velocity is −3.0 m s⁻¹ and its acceleration is +2.0 m s⁻². The acceleration stays constant. Which describes the motion of the body?
Answer and reasoning
Moving left and slowing down; it will then stop and stay at rest. — A student who thinks that slowing down must end in coming to rest picks this. The velocity is zero for only an instant: the acceleration is still +2.0 m s⁻² then, so the velocity keeps changing, becomes positive, and the body moves off to the right.
Moving left and speeding up, since its acceleration is positive. — A student who reads a positive acceleration as 'speeding up' picks this. Whether a body speeds up depends on whether its acceleration and velocity have the same direction. Here they are opposite, so the body is slowing down.
Moving left and slowing down; later it will move off to the right. — The negative velocity means the body is moving left. The acceleration points the opposite way, so the speed decreases: after 1.5 s the velocity is zero. The acceleration is still +2.0 m s⁻² at that instant, so the velocity then becomes positive and the body moves right, speeding up.
Moving right and speeding up, in the direction of its acceleration. — A student who assumes a body moves in the direction of its acceleration picks this. The direction of motion is given by the velocity, −3.0 m s⁻¹, so at this instant the body is moving left.
2 A ball is thrown vertically upwards. Fluid resistance is negligible and upwards is taken as positive. Which statement about the ball at the instant it reaches its highest point is correct?
Answer and reasoning
Its velocity is zero, so its acceleration is zero as well. — A student who thinks zero velocity means zero acceleration picks this. Acceleration is the rate of change of velocity. At the top the velocity is changing from upwards to downwards, so the acceleration is g downwards, the same as everywhere else in the flight.
Its acceleration is reversing from upwards to downwards. — A student who thinks acceleration points in the direction of motion expects an upward acceleration while the ball rises, reversing at the top. The acceleration is g downwards throughout; on the way up it is opposite to the velocity, which is why the ball slows.
Its acceleration is changing sign from negative to positive. — A student who reads negative acceleration as 'slowing down' and positive as 'speeding up' expects the sign to change when the ball starts to speed up on the way down. With upwards positive the acceleration is −g throughout; the ball speeds up on the way down because its velocity is then negative too.
Its velocity is zero; its acceleration is g downwards. — At the top the velocity changes from positive to negative, so it is momentarily zero. It is still changing at that instant, so the acceleration is not zero: it is g downwards (−g), as at every other point of the flight.
3 A runner moves along a straight track. Her position x is measured from a marker on the track. She starts at x = +30 m, runs to x = +70 m, then turns round and runs to x = −20 m, where she stops. What is her displacement?
Answer and reasoning
130 m — A student who equates displacement with distance adds the two legs, 40 m + 90 m = 130 m. That is the distance travelled; displacement depends only on where she started and where she finished.
−50 m — Displacement is the change in position: x(final) − x(initial) = (−20 m) − (+30 m) = −50 m. The turning point at +70 m affects the distance travelled but not the displacement.
−20 m — A student who confuses position with displacement gives her final position, −20 m, measured from the marker. Displacement is measured from where she started, x = +30 m.
+50 m — A student who treats displacement as a length, which cannot be negative, drops the sign. She finishes 50 m on the negative side of her starting point, so the displacement is −50 m; the sign gives its direction.
Working Displacement = x(final) − x(initial) = (−20 m) − (+30 m) = −50 m, i.e. 50 m in the negative direction. (Distance travelled = (70 − 30) m + (70 − (−20)) m = 40 m + 90 m = 130 m.)
4 A body moves from point P to point Q. Which statement about its displacement is correct?
Answer and reasoning
Its magnitude can be less than the distance travelled, but never greater. — Displacement is the straight-line change in position from P to Q; distance is the length of the path. The straight line is the shortest route, so the magnitude of the displacement equals the distance only for one-way motion along a straight line and is otherwise smaller.
It equals the total length of the path followed from P to Q. — A student who treats displacement and distance as the same picks this. The total length of the path is the distance. Displacement depends only on the positions of P and Q.
It depends on the route taken, whereas the distance does not. — A student who uses 'distance' in its everyday sense, the straight-line separation of two places, swaps the two meanings. In physics the distance depends on the route and the displacement does not.
It is always positive, because it is a length measured in metres. — A student who treats displacement as a length picks this. Displacement is a vector; in one dimension it is negative when the change in position is in the negative direction.
5 A car travels 3.0 km along a straight road at a constant speed of 10 m s⁻¹, then returns along the same road to its starting point at a constant speed of 15 m s⁻¹. What are its average speed and its average velocity for the whole journey?
Answer and reasoning
Average speed 12.5 m s⁻¹; average velocity 0 m s⁻¹ — A student who averages the two speeds, (10 + 15)/2, assumes equal times at each speed. The car spends 300 s at 10 m s⁻¹ but only 200 s at 15 m s⁻¹, so average speed = total distance ÷ total time = 12 m s⁻¹.
Average speed 12 m s⁻¹; average velocity 0 m s⁻¹ — The outward leg takes 300 s and the return leg 200 s, so average speed = 6000 m ÷ 500 s = 12 m s⁻¹. The car finishes where it started, so its displacement is zero and so is its average velocity.
Average speed 12 m s⁻¹; average velocity 12 m s⁻¹ — A student who treats displacement as the distance travelled divides 6000 m by 500 s for the velocity as well. The car returns to its start, so its displacement, and therefore its average velocity, is zero.
Average speed 12 m s⁻¹; average velocity −2.5 m s⁻¹ — A student who takes the average velocity as (u + v)/2 averages the outward velocity, +10 m s⁻¹, and the return velocity, −15 m s⁻¹. That shortcut holds only for uniform acceleration. Average velocity is displacement ÷ time, and the displacement for the round trip is zero.
Working Time out = 3000 m ÷ 10 m s⁻¹ = 300 s; time back = 3000 m ÷ 15 m s⁻¹ = 200 s; total time = 500 s. Average speed = total distance ÷ total time = 6000 m ÷ 500 s = 12 m s⁻¹. The car ends at its starting point, so displacement = 0 and average velocity = 0 ÷ 500 s = 0 m s⁻¹.
6 Which statement about when the equations of motion s = ((u + v)/2)t, v = u + at, s = ut + ½at² and v² = u² + 2as can be applied is correct?
Answer and reasoning
They apply to any straight-line motion, even if the acceleration varies. — A student who remembers 'straight line' as the condition picks this. The equations are derived for constant acceleration; if the acceleration varies they give wrong answers even for straight-line motion.
They apply to a skydiver falling through air, as g is constant. — A student who thinks fluid resistance only lowers speed, leaving the acceleration at g, picks this. The fluid resistance on a skydiver grows with speed, so the acceleration decreases from g towards zero as terminal speed is approached; it is not constant.
They apply whenever the acceleration is constant in size and direction. — The equations are derived by assuming constant acceleration, and that is the only condition. They then apply to the whole motion, including through an instant of zero velocity, provided one sign convention is used.
They apply only until the body stops and reverses its direction. — A student who thinks a body that slows to rest must stay there picks this. A constant acceleration carries on through the instant of zero velocity, so one set of equations describes, for example, a ball thrown up and falling back.
7 A graph of velocity against time for a sprinter starts at the origin, rises steeply, becomes gradually less steep, and is a horizontal line at 11 m s⁻¹ from t = 6.0 s until the end of the race. Which describes the sprinter's motion?
Answer and reasoning
The acceleration decreases, so the sprinter is slowing down after the start. — A student who reads decreasing acceleration as slowing down picks this. The velocity rises throughout the first 6.0 s; the acceleration is decreasing but is still forwards, so the sprinter keeps getting faster.
The acceleration increases, since the sprinter keeps getting faster. — A student who links acceleration to speed expects the acceleration to grow as the speed grows. Acceleration is the gradient, not the height, of the graph, and the gradient gets smaller as the curve flattens.
The acceleration falls to zero, and the sprinter is at rest after 6.0 s. — A student who reads any horizontal line on a motion graph as 'at rest' picks this. On a velocity–time graph a horizontal line at 11 m s⁻¹ means a constant velocity of 11 m s⁻¹; the sprinter is running steadily.
The acceleration falls to zero while the speed rises to a steady value. — The gradient of a velocity–time graph is the acceleration. The gradient decreases from a large value to zero at t = 6.0 s, while the height, the velocity, rises to 11 m s⁻¹ and then stays constant.
8 Ball X, of mass 0.20 kg, rolls off a horizontal table top at 3.0 m s⁻¹. At the same instant ball Y, of mass 0.40 kg, is released from rest at the same height. Fluid resistance is negligible. Which ball reaches the floor first?
Answer and reasoning
Y lands first, as X's horizontal motion keeps it in the air longer. — A student who thinks horizontal motion holds a body up picks this. Without fluid resistance, horizontal motion has no effect on the vertical motion; the time to fall depends only on the height and g.
Y lands first, as a heavier ball falls with greater acceleration. — A student who believes heavier bodies fall faster picks this. Without fluid resistance every body falls with the same acceleration, g, whatever its mass.
X lands first, as its greater speed carries it down faster. — A student who does not separate the components treats X's 3.0 m s⁻¹ as helping it downwards. That velocity is horizontal; X's initial vertical velocity is zero, the same as Y's.
They land together, as both start with zero vertical velocity. — The horizontal and vertical motions are independent. Both balls start with zero vertical velocity, fall the same height and have the same vertical acceleration, g, so they take the same time. X's horizontal velocity only decides where it lands.
9 A small steel ball is released from rest in air and falls until it reaches terminal speed. How does its acceleration change during this time?
Answer and reasoning
It is constant but less than g, because fluid resistance is constant. — A student who treats fluid resistance as a fixed force picks this. Fluid resistance depends on speed: it is zero at release and grows as the ball speeds up, so the acceleration does not stay constant.
It stays equal to g, since fluid resistance only lowers the ball's speed. — A student who thinks fluid resistance just slows a body, leaving its acceleration at g, picks this. A lower rate of gain of speed IS a smaller acceleration; if the acceleration stayed at g the ball would never reach a terminal speed.
It increases, since the ball moves faster and faster as it falls. — A student who links acceleration to speed picks this. The ball's speed does increase, but ever more slowly, so its acceleration decreases towards zero.
It decreases from g towards zero as the fluid resistance increases. — At release the ball is at rest, so there is no fluid resistance and its acceleration is g. As it speeds up the fluid resistance grows, so its acceleration decreases, reaching zero at terminal speed.
10 The position x of a body moving along a straight line is recorded at intervals of 1.0 s, starting at t = 0: x = 0, 5.0 m, 8.0 m, 9.0 m, 8.0 m, 5.0 m and 0. Which statement about the motion of the body is correct?
Answer and reasoning
The acceleration is zero at t = 3.0 s, because the body is momentarily at rest at that time. — A student who thinks zero velocity means zero acceleration picks this. The velocity is zero for only an instant at t = 3.0 s, and it is still changing at the same rate (−2.0 m s⁻¹ each second) through that instant, so the acceleration there is −2.0 m s⁻², not zero.
The acceleration is negative up to t = 3.0 s, then positive as the body speeds up again. — A student who reads 'speeding up' as a positive acceleration picks this. The changes in position fall by the same 2.0 m each second throughout, so the acceleration stays −2.0 m s⁻². After t = 3.0 s the velocity is negative and the acceleration is negative, so the body speeds up in the negative direction.
The acceleration is constant and negative; the body reverses direction at t = 3.0 s. — The change in position each second is +5.0 m, +3.0 m, +1.0 m, −1.0 m, −3.0 m, −5.0 m: the velocity falls by 2.0 m s⁻¹ every second, so the acceleration is a constant −2.0 m s⁻². The velocity changes sign between the 3rd and 4th second, at t = 3.0 s, where x is greatest, so the body reverses direction there.
The acceleration decreases while the body slows down, then increases as it speeds up. — A student who links the size of the acceleration to the speed picks this. The speed falls and then rises, but the velocity changes by the same −2.0 m s⁻¹ in every second, so the acceleration is constant throughout.
Working Changes in position over successive 1.0 s intervals: +5.0, +3.0, +1.0, −1.0, −3.0, −5.0 m, so the average velocities are +5.0, +3.0, +1.0, −1.0, −3.0, −5.0 m s⁻¹. Each is 2.0 m s⁻¹ less than the one before, so the acceleration is constant at −2.0 m s⁻². The velocity passes through zero at t = 3.0 s (where x = 9.0 m is greatest): the body reverses direction there while its acceleration stays −2.0 m s⁻².
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
21 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A graph of velocity against time for a trolley moving along a straight track is a straight line from (0 s, 1.0 m s⁻¹) to (2.0 s, 4.0 m s⁻¹). What is the acceleration of the trolley at t = 2.0 s?
Answer and reasoning
1.5 m s⁻² — Acceleration is the rate of change of velocity, the gradient of the velocity–time graph. The line is straight, so the gradient is the same at every instant: (4.0 − 1.0) m s⁻¹ ÷ 2.0 s = 1.5 m s⁻².
2.0 m s⁻² — A student who calculates acceleration as velocity ÷ time divides 4.0 m s⁻¹ by 2.0 s. That works only for a body starting from rest. This trolley started at 1.0 m s⁻¹, so its change in velocity is 3.0 m s⁻¹.
5.0 m s⁻² — A student who confuses area with gradient calculates the area under the line, ½(1.0 + 4.0) × 2.0 = 5.0. That area is the displacement, 5.0 m, not the acceleration; acceleration is the gradient.
4.0 m s⁻² — A student who reads the height of the graph instead of its gradient takes the velocity at t = 2.0 s, 4.0 m s⁻¹. A value read from a velocity–time graph is a velocity; the acceleration is how steeply that value changes.
Working Acceleration = gradient of the velocity–time graph. The graph is a straight line, so the acceleration is uniform and equals the gradient at every instant, including t = 2.0 s: a = Δv/Δt = (4.0 − 1.0) m s⁻¹ / (2.0 − 0) s = 3.0 m s⁻¹ / 2.0 s = 1.5 m s⁻².
2 A car starts from rest at t = 0 and x = 0 and moves along a straight road. A graph of its position x against time t is a curve that gets steeper with time. At t = 3.0 s, x = 16.5 m, and the tangent to the curve at this point also passes through the point (5.0 s, 40.5 m). What is the instantaneous velocity of the car at t = 3.0 s?
Answer and reasoning
5.5 m s⁻¹ — A student who divides position by time, 16.5 m ÷ 3.0 s, finds the AVERAGE velocity over the first 3.0 s (the gradient of the chord from the origin). The car is accelerating, so its velocity at 3.0 s is greater than its average so far.
16.5 m s⁻¹ — A student who reads the height of the graph gives the position at 3.0 s, 16.5 m, as the velocity. A value read from a position–time graph is a position; velocity is its gradient.
0.083 m s⁻¹ — A student who divides the change in time by the change in position calculates 2.0 s ÷ 24 m. A gradient is the change in the vertical-axis quantity (position) divided by the change in the horizontal-axis quantity (time).
12 m s⁻¹ — Instantaneous velocity is the gradient of the tangent to the position–time graph: (40.5 − 16.5) m ÷ (5.0 − 3.0) s = 24 m ÷ 2.0 s = 12 m s⁻¹.
Working Instantaneous velocity = gradient of the tangent to the position–time graph at t = 3.0 s. The tangent passes through (3.0 s, 16.5 m) and (5.0 s, 40.5 m): v = (40.5 − 16.5) m / (5.0 − 3.0) s = 24 m / 2.0 s = 12 m s⁻¹. (For comparison, the average velocity over the first 3.0 s is 16.5 m / 3.0 s = 5.5 m s⁻¹.)
3 A car travelling at 16 m s⁻¹ brakes with uniform deceleration and comes to rest in 4.0 s. What distance does the car travel while braking?
Answer and reasoning
64 m — A student who uses distance = speed × time with the initial speed, 16 × 4.0, treats the car as moving at 16 m s⁻¹ for the whole 4.0 s. The car is slowing, so its average speed, 8.0 m s⁻¹, must be used.
56 m — A student who does not square the time uses s = ut + ½at with a = −4.0 m s⁻²: 64 − ½ × 4.0 × 4.0 = 56. The last term is ½at² = ½ × (−4.0) × (4.0)² = −32 m.
32 m — With uniform acceleration, s = ((u + v)/2)t = ((16 + 0)/2) × 4.0 = 32 m. The same answer follows from s = ut + ½at² with a = −4.0 m s⁻²: 64 − 32 = 32 m.
96 m — A student who enters the magnitude of the deceleration as a = +4.0 m s⁻² gets 64 + 32 = 96. With the car's direction of motion positive, a braking acceleration is negative, −4.0 m s⁻².
Working u = 16 m s⁻¹, v = 0, t = 4.0 s, acceleration uniform. s = ((u + v)/2)t = ((16 + 0)/2 m s⁻¹)(4.0 s) = 32 m. Check: a = (v − u)/t = (0 − 16)/4.0 = −4.0 m s⁻²; s = ut + ½at² = (16)(4.0) + ½(−4.0)(4.0)² = 64 − 32 = 32 m.
4 A stone is released from rest and falls 5.00 m. Fluid resistance is negligible. Take g = 9.8 m s⁻². What is the speed of the stone when it has fallen 5.00 m?
Answer and reasoning
98.0 m s⁻¹ — A student who stops at v² = u² + 2as = 98.0 has found the square of the speed, in m² s⁻². Taking the square root gives v = 9.90 m s⁻¹.
9.90 m s⁻¹ — With downwards positive, v² = u² + 2as = 0 + 2 × 9.8 × 5.00 = 98.0 m² s⁻², so v = √98.0 = 9.90 m s⁻¹.
7.00 m s⁻¹ — A student who leaves out the ½ in s = ½gt² finds t = √(5.00/9.8) = 0.714 s and then v = gt = 7.00 m s⁻¹. With the ½, t = √(2 × 5.00/9.8) = 1.01 s and v = 9.90 m s⁻¹.
4.95 m s⁻¹ — A student who divides the distance by the time of fall, 5.00 m ÷ 1.01 s, finds the AVERAGE speed during the fall. The stone accelerates uniformly from rest, so its final speed is twice this average.
Working Take downwards as positive: u = 0, s = 5.00 m, a = 9.8 m s⁻². v² = u² + 2as = 0 + 2(9.8 m s⁻²)(5.00 m) = 98.0 m² s⁻², so v = √98.0 = 9.90 m s⁻¹.
5 A car starts from rest and moves along a straight road. Its acceleration is greatest at the start and decreases continuously, falling quickly at first and then more slowly, but stays positive, until the car reaches 20 m s⁻¹ at t = 10 s. Which gives the distance travelled by the car in these 10 s?
Answer and reasoning
Between 100 m and 200 m, since its speed rises fastest early on — The velocity–time graph rises steeply and then flattens, so it lies above the straight line from (0, 0) to (10 s, 20 m s⁻¹), whose area is 100 m. The velocity stays below 20 m s⁻¹ until t = 10 s, so the area is also less than 20 × 10 = 200 m.
Exactly 100 m, as its average velocity is (0 + 20)/2 m s⁻¹ — A student who uses (u + v)/2 for any motion gets 10 m s⁻¹ × 10 s. That average holds only for uniform acceleration. Here the car gains speed early, so its average velocity is more than 10 m s⁻¹.
Exactly 200 m, as it moves at 20 m s⁻¹ for the whole 10 s — A student who uses distance = speed × time with the final speed treats the car as moving at 20 m s⁻¹ throughout. It started from rest and was slower than 20 m s⁻¹ until t = 10 s, so it travelled less than 200 m.
Less than 100 m, as the curve of its velocity–time graph bends upwards — A student who copies the shape of the acceleration–time graph, which falls steeply and then levels off (bending upwards), onto the velocity–time graph draws a curve that rises slowly at first and lies below the straight line to (10 s, 20 m s⁻¹), giving less than 100 m. The acceleration is the gradient of the velocity–time graph: it is greatest at the start, so the graph is steepest at first and curves downwards, lying above that line, and the distance exceeds 100 m.
Working Displacement = area under the velocity–time graph. Uniform acceleration from 0 to 20 m s⁻¹ in 10 s would give a straight line with area ½ × 10 s × 20 m s⁻¹ = 100 m. With the acceleration greatest at the start and decreasing, the graph is a curve that rises steeply then flattens, lying above that straight line, so the area exceeds 100 m. The acceleration is positive throughout, so the velocity is below 20 m s⁻¹ before t = 10 s and the area is less than 20 m s⁻¹ × 10 s = 200 m. So 100 m < distance < 200 m.
6 A ball is kicked horizontally at 12 m s⁻¹ from the top of a vertical cliff 20 m high and lands on level ground below. Fluid resistance is negligible. Take g = 9.8 m s⁻². How far from the foot of the cliff does the ball land?
Answer and reasoning
14 m — A student who puts the launch speed into the vertical equation solves 20 = 12t + 4.9t², giving t = 1.14 s and 14 m. The ball is kicked horizontally, so its initial vertical velocity is zero; the 12 m s⁻¹ belongs only to the horizontal motion.
24 m — Vertically, the ball starts with zero vertical velocity: 20 = ½ × 9.8 × t², so t = 2.02 s. Horizontally, its velocity stays 12 m s⁻¹, so it travels 12 × 2.02 = 24 m.
17 m — A student who leaves out the ½ writes 20 = 9.8t², giving t = 1.43 s and 17 m. The vertical displacement from rest is ½gt², so t = √(2 × 20/9.8) = 2.02 s.
49 m — A student who forgets the square root takes t = 2 × 20/9.8 = 4.08 and gets 49 m. That number is t² (in s²); the time of fall is its square root, 2.02 s.
Working Take downwards as positive. Vertical: u_y = 0, s = 20 m, a = 9.8 m s⁻²; s = ½at² gives t = √(2s/a) = √(40/9.8) = 2.02 s. Horizontal: u_x = 12 m s⁻¹ constant (no horizontal acceleration), so x = u_x t = (12 m s⁻¹)(2.02 s) = 24 m.
7 A ball is kicked from level ground at 20.0 m s⁻¹ at 30.0° above the horizontal. Fluid resistance is negligible. Take g = 9.8 m s⁻². What is the maximum height reached by the ball?
Answer and reasoning
15.3 m — A student who swaps sin and cos uses 20.0 cos 30.0° = 17.3 m s⁻¹ as the vertical component. With θ measured from the horizontal, the vertical component is the one opposite θ, u sin θ.
20.4 m — A student who does not resolve uses the launch speed, 20.0 m s⁻¹, in the vertical equation. Only the vertical component, 10.0 m s⁻¹, is reduced to zero by gravity; the horizontal component is unaffected.
5.10 m — The vertical component of the launch velocity is 20.0 sin 30.0° = 10.0 m s⁻¹. At the maximum height the vertical velocity is zero, so 0 = 10.0² − 2 × 9.8 × h, giving h = 5.10 m.
10.2 m — A student who multiplies the initial vertical speed, 10.0 m s⁻¹, by the time to the top, 1.02 s, treats the vertical speed as constant. It falls steadily to zero, so the average vertical speed, 5.0 m s⁻¹, must be used.
Working Vertical component of launch velocity: u_y = u sin θ = 20.0 × sin 30.0° = 10.0 m s⁻¹ (upwards positive). At maximum height v_y = 0: v_y² = u_y² − 2gh, so h = u_y²/(2g) = (10.0 m s⁻¹)² / (2 × 9.8 m s⁻²) = 5.10 m.
8 A stone is thrown from the top of a cliff at 15.0 m s⁻¹ at 30.0° below the horizontal. It lands in the sea 40.0 m below the point of release. Fluid resistance is negligible. Take g = 9.8 m s⁻². How long after release does the stone reach the sea?
Answer and reasoning
2.19 s — The initial vertical velocity is 15.0 sin 30.0° = 7.50 m s⁻¹ downwards. With downwards positive, 40.0 = 7.50t + 4.9t², whose positive root is t = 2.19 s.
2.86 s — A student who treats the stone as launched horizontally sets the initial vertical velocity to zero: 40.0 = 4.9t², t = 2.86 s. A launch below the horizontal has a downward vertical component, 7.50 m s⁻¹, which shortens the fall.
1.82 s — A student who swaps sin and cos uses 15.0 cos 30.0° = 13.0 m s⁻¹ as the vertical component, giving 1.82 s. With θ measured from the horizontal, the vertical component is u sin θ = 7.50 m s⁻¹.
3.72 s — A student who enters the launch component as +7.50 m s⁻¹ while taking upwards as positive (with s = −40.0 m and a = −9.8 m s⁻²) treats the stone as if it were thrown upwards, giving 3.72 s. The component is downwards, so with upwards positive it is −7.50 m s⁻¹.
Working Take downwards as positive. u_y = 15.0 × sin 30.0° = 7.50 m s⁻¹, s = 40.0 m, a = 9.8 m s⁻². s = u_y t + ½at²: 40.0 = 7.50t + 4.9t², i.e. 4.9t² + 7.50t − 40.0 = 0. t = [−7.50 + √(7.50² + 4 × 4.9 × 40.0)] / (2 × 4.9) = (−7.50 + 28.99)/9.8 = 2.19 s (the other root is negative).
9 A ball is launched at 40° above the horizontal over level ground. Fluid resistance is negligible. Which statement describes the ball at its highest point?
Answer and reasoning
Its velocity is momentarily zero; its acceleration is g downwards. — A student who carries over the vertical throw, where the ball stops at the top, picks this. For a launch at an angle only the vertical component is zero at the top; the ball is still moving horizontally.
Its velocity is horizontal and non-zero; its acceleration is g downwards. — At the highest point the vertical component of velocity is zero, but the horizontal component, u cos 40°, is unchanged, so the velocity is horizontal. The only acceleration throughout the flight is g downwards.
Its velocity is horizontal and non-zero; it has no acceleration. — A student who thinks zero vertical velocity means zero acceleration picks this. The vertical velocity is changing from upwards to downwards at the top, so the acceleration there is g downwards.
Its horizontal velocity is less than at launch; its acceleration is g downwards. — A student who thinks the ball's 'forward motion' is used up during flight picks this. With no fluid resistance there is no horizontal acceleration, so the horizontal velocity at the top equals the horizontal velocity at launch.
10 A shuttlecock is hit from ground level at 45° above the horizontal. Video analysis shows that it reaches its highest point 0.58 s after launch, at a horizontal distance of 3.6 m from the launch point, and lands 1.36 s after launch, 5.9 m from the launch point. Which conclusion do these data support?
Answer and reasoning
The data must be faulty, as a projectile's path is symmetrical about its highest point. — A student who expects a symmetrical parabola in every case rejects the data. That symmetry holds only when fluid resistance is negligible, which is not so for a shuttlecock; with fluid resistance the descent is steeper and slower.
Fluid resistance cannot explain this, as it only makes the shuttlecock move more slowly. — A student who thinks fluid resistance just lowers speed without changing the path picks this. Fluid resistance changes the acceleration at every point, which changes the shape of the path and makes it lopsided.
Fluid resistance is significant: the descent is steeper and takes longer than the ascent. — The descent takes 1.36 − 0.58 = 0.78 s, longer than the 0.58 s ascent, and covers only 5.9 − 3.6 = 2.3 m horizontally against 3.6 m on the way up. This lopsided path is what fluid resistance produces: it reduces the horizontal velocity throughout and limits the speed of the fall.
The shuttlecock falls slowly because it is light, not because of fluid resistance. — A student who believes lighter bodies fall more slowly picks this. Without fluid resistance a light body falls with the same acceleration g as a heavy one; the shuttlecock's slow descent is caused by fluid resistance, which is large compared with its small weight.
11 A small plastic ball is launched without spin from level ground at 30° above the horizontal. The motion is compared with an identical launch in which fluid resistance is negligible. Which describes the effect of fluid resistance on the ball's motion?
Answer and reasoning
The range is shorter, and the horizontal velocity component decreases during the flight. — Fluid resistance acts opposite to the velocity at every point. While the ball moves forwards this force has a backward horizontal component, so the horizontal velocity component decreases throughout the flight; with less horizontal speed and a lower, shorter flight, the ball lands closer to the launch point.
The range is shorter, but the horizontal velocity component stays constant in flight. — A student who pictures fluid resistance as acting only upwards, as in terminal-speed diagrams, keeps the horizontal velocity constant. Fluid resistance points opposite to the velocity, which here has a forward component, so the force has a backward horizontal component that reduces the horizontal velocity.
The range is shorter, and the highest point is still reached at exactly half the range. — A student who pictures fluid resistance as shrinking the same symmetrical parabola places the highest point halfway. Because the horizontal velocity keeps falling, the ball covers less horizontal distance on the way down than on the way up, so the highest point lies more than halfway along the range.
The range is unchanged, as the ball follows the same path but moves more slowly. — A student who thinks fluid resistance only slows a body, leaving its path unchanged, picks this. Fluid resistance is a force that changes the acceleration at every point, so the path changes: the ball rises less high and lands closer to the launch point.
12 A runner moves along a straight track. Observer A measures her position from a marker M, taking the direction in which she first runs as positive. Observer B measures her position from a marker N, which is 15 m from M in the positive direction, and takes the same direction as positive. According to observer A, the runner starts at +25 m, runs to +95 m, turns round, runs back to +5.0 m and stops. What is the displacement of the runner according to observer B?
Answer and reasoning
−10 m — A student who takes displacement to be the final position measured from the origin gives the runner's final position according to B, (5.0 − 15) m = −10 m. Displacement is the change in position, final minus initial, and the runner started at +10 m according to B.
+20 m — A student who treats displacement as a length that cannot be negative drops the sign. The runner finishes 20 m on the negative side of where she started, so her displacement is −20 m whichever observer measures it.
−20 m — Displacement is the change in position. According to B the runner starts at +10 m and finishes at −10 m, so her displacement is (−10) − (+10) = −20 m, the same as A finds: moving the origin shifts every position by the same amount and leaves the change unaltered.
160 m — A student who equates displacement with distance adds the two legs: 70 m out and 90 m back. That is the distance travelled; the displacement depends only on the start and finish positions.
Working According to B, positions are 15 m less than according to A: the runner starts at (25 − 15) m = +10 m and finishes at (5.0 − 15) m = −10 m. Displacement = x(final) − x(initial) = (−10 m) − (+10 m) = −20 m. This equals A's value, (5.0 − 25) m = −20 m: displacement is a change in position, so it does not depend on where the origin is placed. (Distance travelled = 70 m + 90 m = 160 m.)
13 A ball is thrown vertically upwards from a point 1.20 m above the ground. It rises to a maximum height of 6.70 m above the ground, falls, and is caught at a point 1.70 m above the ground. Upwards is taken as positive. What are the distance travelled by the ball and its displacement, from the throw to the catch?
Answer and reasoning
Distance 10.5 m; displacement +10.5 m — A student who takes displacement and distance to be the same quantity gives 10.5 m for both. The path length is 10.5 m, but the ball finishes only 0.50 m above where it started, so its displacement is +0.50 m.
Distance 10.5 m; displacement +1.70 m — A student who confuses displacement with position gives the final height above the ground, 1.70 m. Displacement is measured from the starting position, 1.20 m above the ground, not from the ground.
Distance 0.50 m; displacement +10.5 m — A student who uses 'distance' in its everyday sense, the separation of the start and finish points, and 'displacement' for the length of the path, has the two quantities the wrong way round. In physics the distance is the path length, 10.5 m, and the displacement is the change in position, +0.50 m.
Distance 10.5 m; displacement +0.50 m — Distance is the length of the path: 5.50 m up then 5.00 m down, 10.5 m in total. Displacement is the change in position, final minus initial: 1.70 m − 1.20 m = +0.50 m, upwards.
Working Distance travelled = length of path = (6.70 − 1.20) m up + (6.70 − 1.70) m down = 5.50 m + 5.00 m = 10.5 m. Displacement = change in position = (1.70 − 1.20) m = +0.50 m (0.50 m upwards).
14 The graph shows how the velocity of a car varies with time during the first 10 s of its motion along a straight road. What is the displacement of the car during these 10 s?
Answer and reasoning
200 m — A student who uses distance = speed × time with the final speed, 20 m s⁻¹ × 10 s, picks this. The car is only at 20 m s⁻¹ from 4.0 s onwards; during the first 4.0 s it is slower, so the displacement is less than 200 m.
100 m — A student who takes the average velocity to be (u + v)/2 = (0 + 20)/2 = 10 m s⁻¹ for the whole 10 s picks this. That formula applies only to uniform acceleration; here the acceleration is not uniform over the 10 s, and the area under the graph must be used.
160 m — Displacement is the area between the velocity–time graph and the time axis: a triangle of area ½ × 4.0 × 20 = 40 m for the first 4.0 s, plus a rectangle of area 6.0 × 20 = 120 m for the next 6.0 s, giving 160 m.
2.0 m — A student who confuses the two graph operations finds the gradient, (20 − 0)/(10 − 0) = 2.0, and reports it as a displacement. The gradient of a velocity–time graph is an acceleration (units m s⁻²); the displacement is the area under the graph.
Working Displacement = area under the velocity–time graph. Triangle from 0 to 4.0 s: ½ × 4.0 s × 20 m s⁻¹ = 40 m. Rectangle from 4.0 s to 10 s: 6.0 s × 20 m s⁻¹ = 120 m. Total displacement = 40 m + 120 m = 160 m.
15 The graph shows how the position x of a body moving along a straight line varies with time t. O, A, B and C are points on the graph. Which statement about the motion is correct?
Answer and reasoning
During BC the body slows down, because the line slopes downwards. — A student who reads the graph as a picture of the motion picks this. BC is a straight line, so its gradient, the velocity, is constant at −2 m s⁻¹: the body moves at a steady 2 m s⁻¹ in the negative direction. Only a curved section would show a changing speed.
During AB the body moves at a constant velocity of 6 m s⁻¹. — A student who reads the height of the graph as the velocity picks this. The height of a position–time graph is the position: during AB the position stays at 6 m, so the gradient, and hence the velocity, is zero and the body is at rest.
During BC the body moves at constant velocity in the negative direction. — Velocity is the gradient of a position–time graph. BC is a straight line, so its gradient is constant; the gradient is negative, (−2 − 6) m ÷ (8 − 4) s = −2 m s⁻¹, so the body moves steadily in the negative direction, passing through x = 0 at t = 7 s.
During OA the acceleration is greatest, because the body moves fastest then. — A student who links a large speed to a large acceleration picks this. During OA the body does move fastest, at 3 m s⁻¹, but OA is a straight line, so the velocity is constant and the acceleration is zero.
16 The graph shows how the velocity of a body moving along a straight line varies with time. The dashed line is the tangent to the curve at P, where t = 2.0 s. What is the instantaneous acceleration of the body at t = 2.0 s?
Answer and reasoning
2.0 m s⁻² — The instantaneous acceleration is the gradient of the tangent at P. Reading the two marked points on the tangent, (12 − 2.0) m s⁻¹ ÷ (5.0 − 0) s = 2.0 m s⁻².
3.0 m s⁻² — A student who calculates a = v/t with the velocity and time at P, 6.0 m s⁻¹ ÷ 2.0 s, picks this. That is the gradient of the chord from the origin to P, the AVERAGE acceleration over the first 2.0 s. The curve is flattening, so the acceleration at t = 2.0 s is less than that average.
6.0 m s⁻² — A student who reads the height of the graph at P as the acceleration picks this. The height of a velocity–time graph is a velocity, 6.0 m s⁻¹; the acceleration is the gradient of the tangent.
0.5 m s⁻² — A student who divides the change in time by the change in velocity, 5.0 s ÷ 10 m s⁻¹, picks this. A gradient is the change in the vertical-axis quantity divided by the change in the horizontal-axis quantity: Δv/Δt, giving units m s⁻².
Working Instantaneous acceleration = gradient of the tangent to the velocity–time graph at P. The tangent passes through (0, 2.0 m s⁻¹) and (5.0 s, 12 m s⁻¹): a = (12 − 2.0) m s⁻¹ ÷ (5.0 − 0) s = 10 m s⁻¹ ÷ 5.0 s = 2.0 m s⁻². (For comparison, the average acceleration over the first 2.0 s is 6.0 m s⁻¹ ÷ 2.0 s = 3.0 m s⁻².)
17 The graph shows how the velocity of a body moving along a straight line varies with time. What is the displacement of the body between t = 0 and t = 4.0 s?
Answer and reasoning
16 m — Read u = +12 m s⁻¹ and v = −4.0 m s⁻¹ at t = 4.0 s from the graph. With uniform acceleration, s = ((u + v)/2)t = (12 − 4.0)/2 × 4.0 = 16 m. Equivalently, the area above the axis (+18 m) plus the area below it (−2.0 m) gives +16 m.
20 m — A student who treats displacement as the distance travelled adds the two areas without regard to sign: 18 m + 2.0 m = 20 m. After t = 3.0 s the velocity is negative, so the body moves back and that 2.0 m reduces the displacement.
32 m — A student who enters the final velocity as +4.0 m s⁻¹, ignoring its sign, calculates ((12 + 4.0)/2) × 4.0 = 32 m. The graph shows v = −4.0 m s⁻¹ at 4.0 s; the sign must be carried into the equation.
40 m — A student who finds a = −4.0 m s⁻² correctly but uses s = ut + ½at without squaring the time gets 48 − 8 = 40 m. With t² the second term is ½ × (−4.0) × 16 = −32 m, giving 16 m.
Working From the graph, u = +12 m s⁻¹, v = −4.0 m s⁻¹ at t = 4.0 s. The line is straight, so the acceleration is uniform and s = ((u + v)/2)t = ((12 + (−4.0))/2) × 4.0 = 4.0 m s⁻¹ × 4.0 s = 16 m. Check by areas: +½ × 3.0 × 12 = +18 m above the axis, −½ × 1.0 × 4.0 = −2.0 m below it, total +16 m. (a = (−4.0 − 12)/4.0 = −4.0 m s⁻²; s = ut + ½at² = 48 − 32 = 16 m.)
18 The graph shows how the acceleration of a dragster moving along a straight track varies with time. The dragster starts from rest at t = 0. What is the velocity of the dragster at t = 15 s?
Answer and reasoning
120 m s⁻¹ — A student who applies v = u + at with a = 8.0 m s⁻² for the whole 15 s picks this. That equation needs a constant acceleration; here the acceleration falls to zero after 5.0 s, so the velocity gained after 5.0 s is only half of 8.0 × 10.
0 m s⁻¹ — A student who reads the height of the graph at t = 15 s, which is zero, as the velocity picks this. The height of an acceleration–time graph is the acceleration: zero acceleration at 15 s means the velocity has stopped changing, not that it is zero.
−0.80 m s⁻¹ — A student who confuses gradient with area finds the gradient of the sloping section, (0 − 8.0)/(15 − 5.0) = −0.80, and reports it as a velocity. The gradient of an acceleration–time graph is the rate of change of acceleration; the velocity change is the area under the graph.
80 m s⁻¹ — The area under an acceleration–time graph is the change in velocity: 8.0 × 5.0 = 40 m s⁻¹ for the uniform section, plus ½ × 10 × 8.0 = 40 m s⁻¹ for the section in which the acceleration falls to zero. From rest, v = 80 m s⁻¹.
Working Change in velocity = area under the acceleration–time graph. Rectangle from 0 to 5.0 s: 8.0 m s⁻² × 5.0 s = 40 m s⁻¹. Triangle from 5.0 s to 15 s: ½ × 10 s × 8.0 m s⁻² = 40 m s⁻¹. Starting from rest, v = 40 + 40 = 80 m s⁻¹.
19 The diagram shows the path of a ball launched from O over level ground. Fluid resistance is negligible. P, Q and R are three points on the path. Which statement about the ball is correct?
Answer and reasoning
The horizontal component of velocity at R is less than at P. — A student who imagines the ball's forward motion being used up picks this. With fluid resistance negligible the only force is the weight, which is vertical, so the horizontal component of velocity is the same at P, Q and R.
At P the acceleration of the ball is directed along the path, towards Q. — A student who assumes the acceleration points the way the body is moving picks this. The acceleration is that of free fall, g vertically downwards, at every point of the path, including P where the ball is moving upwards and forwards.
At Q the velocity is zero, as the ball is momentarily at rest. — A student who carries over 'at rest at the top' from a ball thrown straight up picks this. At Q only the vertical component of velocity is zero; the horizontal component is unchanged, so the ball is still moving horizontally at Q.
The speed of the ball at R is equal to the speed of the ball at P. — The horizontal velocity component is constant throughout, and the vertical motion is symmetrical about Q: on returning to the height of P the vertical component has the same magnitude as at P but points downwards. Equal components in magnitude give equal speeds; only the direction of the velocity differs.
20 The graph shows how the velocity of a small ball released from rest varies with time. Curve P is for the ball falling in air. Line Q is for the same ball falling with fluid resistance negligible. Which statement about the graph is correct?
Answer and reasoning
The gradient of P is less than that of Q from t = 0, since fluid resistance acts from release. — A student who treats fluid resistance as a fixed force picks this. Fluid resistance depends on speed and is zero for a ball at rest, so at t = 0 the acceleration is g for both cases and the two graphs start with the same gradient; they separate only as the ball gains speed.
Once P becomes horizontal the ball in air is at rest, as its velocity has stopped changing. — A student who reads a horizontal line on any motion graph as 'at rest' picks this. On a velocity–time graph a horizontal line means a constant velocity: the ball is falling steadily at its terminal speed, with zero acceleration.
The gradient of P increases as the ball in air speeds up, as a faster body accelerates more. — A student who links a larger speed to a larger acceleration picks this. The graph shows P becoming less steep as the speed rises: the fluid resistance increases with speed, so the acceleration decreases towards zero.
At t = 0 the gradient of P equals the gradient of Q, as the fluid resistance is zero at release. — The gradient of a velocity–time graph is the acceleration. At release the ball is at rest, so there is no fluid resistance and its acceleration is g, the same as for Q: P starts tangent to Q. As the ball speeds up, the fluid resistance grows, the acceleration falls and P bends away below Q, becoming horizontal at the terminal speed.
Working Gradient of a velocity–time graph = acceleration. At t = 0 the ball is at rest, so the fluid resistance (which depends on speed) is zero and the acceleration is g for both cases: P and Q start with the same gradient, g. As the ball in air gains speed the fluid resistance grows, the resultant force and acceleration fall, and P curves below Q, becoming horizontal (zero acceleration, constant velocity) at the terminal speed.
21 A ball is thrown vertically upwards in air and falls back. The graph shows how its velocity varies with time, with upwards taken as positive; the ball reaches its highest point at t₁. Which statement about the motion is correct?
Answer and reasoning
The acceleration has a larger magnitude before t₁, as the graph is steeper there. — The gradient of the graph is the acceleration, and the graph is steeper before t₁ than after it. While the ball rises, the fluid resistance acts downwards, with the weight, so the acceleration exceeds g in magnitude; while it falls, the fluid resistance acts upwards, against the weight, so the acceleration is less than g and falls towards zero at the terminal speed.
The acceleration is g throughout, as fluid resistance changes only the speed. — A student who thinks fluid resistance merely makes a body slower, leaving its acceleration at g, picks this. The gradient of the graph is not constant: it is steeper than g while the ball rises and shallower than g while it falls, because the fluid resistance adds to the weight on the way up and opposes it on the way down.
At t₁ the acceleration is zero, since the velocity is zero there. — A student who thinks zero velocity means zero acceleration picks this. At t₁ the graph crosses the axis with a clearly non-zero gradient: the fluid resistance is zero at that instant, so the acceleration is exactly g downwards.
After t₁ the acceleration is positive, because the ball is speeding up. — A student who reads 'speeding up' as a positive acceleration picks this. After t₁ the velocity becomes more and more negative, so the gradient, and hence the acceleration, is negative (downwards) even though the ball's speed is increasing.
That was your twenty minutes. Real practice on A.1 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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