A.1 Kinematics

Practice Worksheet — name: ______________________ date: ____________

Formulas for this topic

Final velocity

v = u + at

Displacement

s = ut + \tfrac{1}{2}at^{2}

Velocity–displacement

v^{2} = u^{2} + 2as

Displacement (average velocity)

s = \frac{(u + v)}{2}\,t

Section A — Multiple choice

A1. A ball is thrown vertically upwards. At the highest point of its motion, which statement is correct?

A Velocity and acceleration are both zero
B Velocity is zero; acceleration is $9.81\ \text{m s}^{-2}$ downwards
C Velocity is zero; acceleration is $9.81\ \text{m s}^{-2}$ upwards
D Velocity and acceleration are both non-zero

A2. A car accelerates uniformly from rest to $24\ \text{m s}^{-1}$ over a distance of $144\ \text{m}$ . What is its acceleration?

A $1.0\ \text{m s}^{-2}$
B $2.0\ \text{m s}^{-2}$
C $3.0\ \text{m s}^{-2}$
D $4.0\ \text{m s}^{-2}$

A3. Two identical balls are launched horizontally from the same height, one at $5\ \text{m s}^{-1}$ and one at $10\ \text{m s}^{-1}$ . Ignoring air resistance, which lands first?

A The slower ball
B The faster ball
C They land at the same time
D It depends on their masses

Section B — Short answer

B1. Distinguish between distance and displacement, using an athlete completing one lap of a 400 m track as your example. [2 marks]

B2. A stone is projected horizontally at $12\ \text{m s}^{-1}$ from a cliff 45 m high. Calculate the time of flight and the horizontal range. [4 marks]

B3. Sketch the velocity–time graph for an object falling through air until it reaches terminal speed, and explain its shape. [3 marks]

Answer key

For worked explanations, interactive practice and more free resources, visit www.newtonine.com

Section A

A1: Velocity is zero; acceleration is $9.81\ \text{m s}^{-2}$ downwards — At the top of the flight the ball is momentarily at rest, but gravity never switches off — the acceleration remains $g$ downwards throughout. This is the single most common misconception in kinematics.

A2: $2.0\ \text{m s}^{-2}$ — No time is given, so use $v^2 = u^2 + 2as$ : $24^2 = 0 + 2a(144)$ , giving $a = 576/288 = 2.0\ \text{m s}^{-2}$ . Choosing the equation that avoids the unknown you don't need is the key skill.

A3: They land at the same time — Vertical and horizontal motion are independent. Both balls start with zero vertical velocity and fall the same height under the same $g$ , so their flight times are identical — the faster one simply lands further away.

Section B

B1: Distance is the scalar length of the path travelled — 400 m for the lap. Displacement is the vector change in position from start to finish — zero for a complete lap, since the athlete returns to the starting point.

B2: Vertical: $45 = \frac{1}{2}(9.81)t^2$ gives $t = \sqrt{9.17} \approx 3.0\ \text{s}$ . Horizontal: constant velocity, so range $= 12 \times 3.0 = 36\ \text{m}$ . The two directions are treated independently, linked only by time.

B3: The graph starts at the origin with gradient $g$ (only weight acts initially). As speed increases, fluid resistance grows, the resultant force and hence gradient (acceleration) decrease, and the curve flattens, approaching a horizontal asymptote at terminal speed where drag balances weight.