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How do we describe and predict the motion of an object?

Displacement, velocity and acceleration; motion graphs; the equations of uniformly accelerated motion (suvat); projectile motion as independent horizontal and vertical components.

A focused answer to the Edexcel 9PH0 kinematics content, covering displacement, velocity and acceleration, motion graphs, the suvat equations of uniformly accelerated motion, and projectile motion as independent horizontal and vertical components.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
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What this dot point is asking

Edexcel wants you to define displacement, velocity and acceleration, interpret and sketch motion graphs, use the suvat equations for uniform acceleration, and analyse projectile motion by treating the horizontal and vertical components independently.

The answer

Displacement, velocity and acceleration

Distance and speed are the scalar partners of displacement and velocity. Because velocity is a vector, an object moving in a circle at constant speed is still accelerating, since its direction (and so its velocity) is changing.

Motion graphs

Reading graphs both ways (gradient and area) is a core skill. A negative gradient on a displacement-time graph means motion in the negative direction; an area below the axis on a velocity-time graph counts as negative displacement.

The suvat equations

List the known quantities, choose the equation that contains your three knowns and the unknown, and solve. These apply only when the acceleration is constant; for free fall near the ground, a=g9.81a = g \approx 9.81 m per second squared downward.

Projectile motion

A projectile experiences a constant downward acceleration gg and no horizontal force (ignoring air resistance). The horizontal and vertical motions are independent: horizontally the velocity is constant (x=uxtx = u_x t); vertically the suvat equations apply with a=ga = g. Combining them gives a parabolic trajectory. For a projectile launched at angle θ\theta with speed uu, resolve into ux=ucosθu_x = u\cos\theta and uy=usinθu_y = u\sin\theta, then treat each direction separately.

Examples in context

Stopping-distance calculations for road safety use v2=u2+2asv^2 = u^2 + 2as to relate speed to braking distance, which is why speed limits matter so much (distance grows with the square of speed). Sports such as basketball, golf and long jump are projectile problems where launch angle and speed set the range. Ballistics, fountain design, and the trajectory of water from a hose all rely on the independence of horizontal and vertical motion.

Try this

Q1. State what the area under a velocity-time graph represents. [1 mark]

  • Cue. The displacement of the object.

Q2. A stone is dropped from rest and falls for 3.03.0 s. Find its speed on impact. Take g=9.81g = 9.81 m per second squared. [2 marks]

  • Cue. v=u+at=0+9.81×3.0=29v = u + at = 0 + 9.81 \times 3.0 = 29 m per second.

Q3. Explain why the horizontal velocity of a projectile is constant (ignoring air resistance). [2 marks]

  • Cue. There is no horizontal force, so by Newton's first law there is no horizontal acceleration and the horizontal velocity stays the same.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20174 marksA car accelerates uniformly from 8.08.0 m per second to 2020 m per second in 6.06.0 s. Calculate its acceleration and the distance travelled.
Show worked answer →

Acceleration: a=vut=208.06.0=2.0a = \frac{v - u}{t} = \frac{20 - 8.0}{6.0} = 2.0 m per second squared.

Distance: s=(u+v)2t=(8.0+20)2×6.0=14×6.0=84s = \frac{(u + v)}{2}t = \frac{(8.0 + 20)}{2} \times 6.0 = 14 \times 6.0 = 84 m.

Markers reward the correct suvat selection, the acceleration 2.02.0 m per second squared, and the distance 8484 m.

Edexcel 20215 marksA ball is thrown horizontally at 1212 m per second from a cliff 2020 m high. Determine the time to reach the ground and the horizontal distance travelled. Take g=9.81g = 9.81 m per second squared.
Show worked answer →

Vertical motion (independent): s=12gt2s = \frac{1}{2}gt^2, so 20=12×9.81×t220 = \frac{1}{2} \times 9.81 \times t^2, giving t2=409.81=4.08t^2 = \frac{40}{9.81} = 4.08 and t=2.0t = 2.0 s.

Horizontal motion (constant velocity): x=uxt=12×2.0=24x = u_x t = 12 \times 2.0 = 24 m.

Markers reward treating the vertical and horizontal motions independently, the fall time 2.02.0 s, and the range 2424 m.

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