PhysicsCore28 min read

Kinematics

Describing motion: displacement, velocity, acceleration and the graphs

This topic appears in:

9th Class Physics · ch.21st Year Physics · ch.2O Level Physics (5054) · ch.1A Level Physics (9702) · ch.2A Level Mathematics (9709) · ch.4
01

Scalars and vectors

Definition

Scalar — A quantity with magnitude only. Distance, speed, time and mass are scalars.

A second definition sits beside it, and the whole chapter depends on keeping the two apart.

ScalarVector equivalentDifference
Distance — total path lengthDisplacement — straight line from start to finishdisplacement has direction and can be zero after a journey
Speed — rate of change of distanceVelocity — rate of change of displacementvelocity changes if direction changes, even at constant speed
Mass, time, energyForce, acceleration, momentum—
Worked example 1

A runner completes one lap of a 400 m circular track in 50 s. Find (a) the average speed and (b) the average velocity.

  1. Average speed = total distance ÷ time = 400 ÷ 50 = 8.0 m s⁻¹.Distance counts the whole path, so a full lap is 400 m.
  2. Displacement = 0, because the finish point is the start point.Displacement measures the straight line between the two, not the route.
  3. Average velocity = displacement ÷ time = 0 ÷ 50.

(a) 8.0 m s⁻¹ (b) 0 m s⁻¹

Read the question for the word "velocity"

A question about a return journey that asks for average velocity is almost always testing whether you noticed the displacement is zero. The word is chosen deliberately. The same applies to "distance" against "displacement".

02

Acceleration

Definition

Acceleration — The rate of change of velocity. Because velocity is a vector, an object accelerates if its speed changes, if its direction changes, or both.

A car going round a roundabout at a steady 30 km/h is accelerating. The speedometer never moves, but the direction does, so the velocity does.

Negative acceleration does not mean "slowing down". It means acceleration in the negative direction. An object moving backwards and speeding up has negative velocity and negative acceleration. To decide whether something is speeding up, compare the signs of velocity and acceleration: same sign speeds up, opposite signs slows down.

VelocityAccelerationThe object is
positivepositivemoving forwards, speeding up
positivenegativemoving forwards, slowing down
negativenegativemoving backwards, speeding up
negativepositivemoving backwards, slowing down
a = Δv / Δt = (v − u) / tdefining equation for uniform acceleration
u
initial velocitym s⁻¹
v
final velocitym s⁻¹
a
accelerationm s⁻²
t
time takens
03

The equations of motion

Four equations connect the five quantities u, v, a, s and t. Each one leaves out exactly one quantity, and choosing the right equation is simply a matter of spotting which quantity the question does not mention.

They are valid only when acceleration is constant. For non-uniform acceleration you need a graph or calculus instead.

v = u + at(no s)s = ut + ½at²(no v)v² = u² + 2as(no t)s = ½(u + v)t(no a)the SUVAT equations — constant acceleration only
s
displacementm
u
initial velocitym s⁻¹
v
final velocitym s⁻¹
a
accelerationm s⁻²
t
times
Worked example 2

A car accelerates uniformly from rest and covers 100 m in 8.0 s. Calculate its acceleration and its final velocity.

  1. List: u = 0, s = 100, t = 8.0, a = ?, v = ?.v is missing from the given data, so start with the equation that omits v.
  2. Use s = ut + ½at²: 100 = 0 + ½ × a × 8.0².u = 0 kills the first term.
  3. 100 = 32a, so a = 3.125 ≈ 3.1 m s⁻².½ × 64 = 32.
  4. Then v = u + at = 0 + 3.125 × 8.0 = 25 m s⁻¹.Use the unrounded value of a here to avoid rounding error.

a = 3.1 m s⁻² v = 25 m s⁻¹

Key points

  1. Write out u, v, a, s, t and fill in what you know before choosing an equation. The gap tells you which one to use.
  2. "From rest" means u = 0. "Comes to rest" means v = 0. Both are marks in disguise.
  3. Take one direction as positive and keep it for the whole question. Downward-positive is usually easiest for falling objects.
  4. Freely falling means a = g ≈ 9.81 m s⁻² (use 10 if the paper says so), downward, whatever the mass.
04

Motion graphs

Two rules cover every motion graph you will ever be given. The gradient gives you the next quantity down. The area under the line gives you the previous quantity up.

Learn those two sentences and you never need to memorise the shape of a particular graph again.

  • A straight line on a displacement–time graph means constant velocity; a curve means the velocity is changing.
  • A horizontal line on a velocity–time graph means constant velocity, so zero acceleration — not a stationary object.
  • Area below the time axis on a velocity–time graph counts as negative displacement. For total distance, add the areas ignoring sign; for displacement, keep the signs.
  • On a curved velocity–time graph, the acceleration at an instant is the gradient of the tangent at that point.
GraphGradient givesArea under gives
Displacement–timevelocitynothing meaningful
Velocity–timeaccelerationdisplacement
Acceleration–timerate of change of accelerationchange in velocity

Finding the gradient properly

Use a large triangle spanning most of the line, and read the coordinates off the axes rather than counting squares. Examiners award a mark for a triangle that is clearly big enough, and take it off for one drawn over two centimetres.

05

Free fall and projectiles

Near the Earth's surface, and ignoring air resistance, every object falls with the same acceleration g ≈ 9.81 m s⁻² downward — regardless of mass. A feather and a hammer dropped on the Moon land together, and the Apollo 15 crew filmed it.

For projectile motion the single most useful idea is this: horizontal and vertical motion are independent. Gravity acts downward, so the vertical motion accelerates. Nothing acts horizontally, so the horizontal velocity is constant. Solve them as two separate columns joined only by the shared time.

horizontal: s_x = u_x t(a_x = 0)vertical:s_y = u_y t + ½gt²two independent problems, linked by t
u_x
horizontal component of initial velocity, = u cos θm s⁻¹
u_y
vertical component of initial velocity, = u sin θm s⁻¹
g
acceleration of free fall9.81 m s⁻² downward

Set angle to 45° for maximum range on level ground. Then compare 30° and 60° — the same range, because sin 2θ takes the same value for both. The 60° shot simply spends longer in the air and goes higher.

Key points

  1. Time of flight is decided entirely by the vertical motion. A ball dropped and a ball fired horizontally from the same height land together.
  2. At the highest point of a projectile's path the vertical velocity is zero, but the horizontal velocity — and therefore the total velocity — is not.
  3. On level ground, range = u² sin 2θ / g, greatest at θ = 45°.
  4. Air resistance shortens the range, lowers the maximum height, and makes the descent steeper than the ascent — the path is no longer a symmetric parabola.

Practice questions

8 questions · 31 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

2 · 4 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Differentiate between distance and displacement.
Model answer

Distance is the total length of the path travelled, a scalar. Displacement is the straight line from start to finish together with its direction, a vector.

Examiner tip. Naming scalar and vector is worth the second mark on its own.

SQ2[2 marks]
Can a body have zero velocity and non-zero acceleration? Explain.
Model answer

Yes. A ball thrown vertically upward is momentarily at rest at the top of its flight, but gravity still acts, so its acceleration is g downward.

Examiner tip. "Yes" with no example scores nothing. The example is the answer.

Solved numericals

1 · 4 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A car travelling at 25 m s⁻¹ brakes uniformly and stops in 5.0 s. Calculate the deceleration and the distance travelled while braking.

Given. u = 25 m s⁻¹, v = 0, t = 5.0 s

Full working
  1. Uses a = (v − u)/t = (0 − 25)/5.0[1]
  2. a = −5.0 m s⁻², a deceleration of 5.0 m s⁻²the negative sign or the word deceleration, not both required[1]
  3. Uses s = ½(u + v)t or v² = u² + 2as[1]
  4. s = ½(25 + 0) × 5.0 = 62.5 m[1]

deceleration 5.0 m s⁻², distance 62.5 m

Exam questions

5 · 23 marks

Multi-part questions with a full mark scheme.

Q1[4 marks]
A cyclist travelling at 4.0 m s⁻¹ accelerates uniformly to 10.0 m s⁻¹ over a distance of 42 m.
  1. Calculate the acceleration of the cyclist.
  2. Calculate the time taken.
Mark scheme
  1. Selects v² = u² + 2asthe equation without t, since t is not given in part (a)[1]
  2. 10.0² = 4.0² + 2 × a × 42 → 100 = 16 + 84a → a = 1.0 m s⁻²unit required for the mark[1]
  3. Selects v = u + at (or s = ½(u+v)t)[1]
  4. 10.0 = 4.0 + 1.0t → t = 6.0 s[1]

(a) 1.0 m s⁻² (b) 6.0 s

Examiner tip. Part (b) can be done from the original data rather than from your answer to (a). If you do use your own answer and (a) was wrong, you can still earn full marks in (b) under error-carried-forward — but only if your working is shown.

Q2[5 marks]
A stone is dropped from rest at the top of a cliff and hits the sea 3.2 s later. Take g = 9.81 m s⁻² and ignore air resistance.
  1. Calculate the height of the cliff.
  2. Calculate the speed at which the stone hits the water.
  3. State one effect of air resistance on your answer to (b).
Mark scheme
  1. Uses s = ut + ½at² with u = 0"dropped from rest" is what tells you u = 0[1]
  2. s = ½ × 9.81 × 3.2² = 50.2 ≈ 50 m[1]
  3. Uses v = u + ator v² = u² + 2as[1]
  4. v = 9.81 × 3.2 = 31.4 ≈ 31 m s⁻¹[1]
  5. The actual speed would be lower / the stone would reach terminal velocitya statement about direction of change is enough[1]

(a) 50 m (b) 31 m s⁻¹ (c) the speed would be less

Examiner tip. Three significant figures is the safe default when the data has three. Writing 50.2 m and 31.4 m s⁻¹ would also be accepted; writing 50.208 m would not, because it claims precision the data does not support.

Q3[6 marks]
The velocity–time graph of a train shows: a uniform rise from 0 to 20 m s⁻¹ over the first 40 s, a constant 20 m s⁻¹ for the next 60 s, then a uniform fall to rest over the final 30 s.
  1. Calculate the acceleration during the first 40 s.
  2. Calculate the total distance travelled.
  3. Calculate the average speed for the whole journey.
Mark scheme
  1. Acceleration = gradient = (20 − 0) / 40gradient of a velocity–time graph is acceleration[1]
  2. = 0.50 m s⁻²[1]
  3. Recognises distance = area under the graphthis is the mark most often missed[1]
  4. Triangle ½ × 40 × 20 = 400; rectangle 60 × 20 = 1200; triangle ½ × 30 × 20 = 300all three areas needed[1]
  5. Total = 1900 m[1]
  6. Average speed = 1900 / 130 = 14.6 ≈ 15 m s⁻¹total distance ÷ total time, not the mean of the velocities[1]

(a) 0.50 m s⁻² (b) 1900 m (c) 15 m s⁻¹

Examiner tip. Average speed is never the average of the two end velocities unless the acceleration is uniform for the whole journey. Here it is not, so it must be total distance over total time.

Q4[5 marks]
A ball is thrown horizontally at 15 m s⁻¹ from the top of a building 45 m high. Take g = 10 m s⁻² and ignore air resistance.
  1. Calculate the time the ball takes to reach the ground.
  2. Calculate the horizontal distance travelled.
  3. Explain why the time in (a) does not depend on the horizontal speed.
Mark scheme
  1. Uses vertical motion with u_y = 0: 45 = ½ × 10 × t²"thrown horizontally" means the initial vertical velocity is zero[1]
  2. t² = 9.0, t = 3.0 s[1]
  3. Uses s_x = u_x t with constant horizontal velocity[1]
  4. s_x = 15 × 3.0 = 45 m[1]
  5. Horizontal and vertical motion are independent / gravity acts only vertically, so the vertical motion is unaffected by the horizontal velocity[1]

(a) 3.0 s (b) 45 m (c) the two components are independent

Examiner tip. Part (c) is worth as much as a calculation and takes ten seconds. Explain-marks are the cheapest marks on any physics paper and the most commonly left blank.

Q5[3 marks]
Define displacement, and state one situation in which the magnitude of an object's displacement is smaller than the distance it has travelled.
Mark scheme
  1. Displacement is the straight-line distance from the starting point to the finishing point[1]
  2. …together with its direction / it is a vector quantitythe direction is required for the second mark[1]
  3. Any curved or non-straight path, e.g. a runner going round a bend, a car following a winding roada full circular lap, where displacement is zero, also earns this[1]

Examiner tip. Definition questions are marked point by point. "Distance in a straight line" alone scores 1 of 2 — the vector nature is a separate mark, so always add it.