Worked Solutions
Module 1: Kinematics — Worked Solutions (Preliminary Physics)
Created with Intu AI Reviewed by Intuition's expert tutors
Studying this? See our Preliminary Physics course →
Worked examples for Preliminary Physics Module 1: Kinematics. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
For motion problems, always start by listing what you know and choosing a positive direction. Take $g = 9.8\ \text{m s}^{-2}$ unless told otherwise.
Example 1 — Vertical launch and equations of motion
Question
A ball is thrown vertically upward from ground level with an initial speed of $14.7\ \text{m s}^{-1}$. Taking $g = 9.8\ \text{m s}^{-2}$ and ignoring air resistance, find the maximum height reached and the total time the ball is in the air before returning to the ground.
Solution
Take up as positive, so $a = -9.8\ \text{m s}^{-2}$ and $u = 14.7\ \text{m s}^{-1}$.
At the top, $v = 0$. Use $v^2 = u^2 + 2as$:
$0 = (14.7)^2 + 2(-9.8)s \Rightarrow s = \dfrac{216.09}{19.6} = 11.025\ \text{m}$.
So maximum height is $\approx 11.0\ \text{m}$.
Time up: $v = u + at \Rightarrow 0 = 14.7 - 9.8t \Rightarrow t = 1.5\ \text{s}$. By symmetry the flight is twice that, so total time $= 3.0\ \text{s}$.
Don't compute the down-leg separately when symmetry gives it to you for free.
Let's set a sign convention first — that's what stops the signs tripping you up. Call up positive, so the initial velocity is $u = +14.7\ \text{m s}^{-1}$ and gravity pulls down, giving $a = -9.8\ \text{m s}^{-2}$.
At the highest point the ball is momentarily still, so $v = 0$. The equation linking velocity and displacement without time is $v^2 = u^2 + 2as$:
$0 = (14.7)^2 + 2(-9.8)s$, so $s = \dfrac{216.09}{19.6} = 11.025\ \text{m} \approx 11.0\ \text{m}$.
For the time, use $v = u + at$: $0 = 14.7 - 9.8t$, giving $t = 1.5\ \text{s}$ to reach the top. Because the path up and down are mirror images (same start and end height), the ball takes the same time coming back, so the total time aloft is $2 \times 1.5 = 3.0\ \text{s}$.
The symmetry trick works because the only force is constant gravity, so the motion is identical in reverse.
Up positive: $u = 14.7\ \text{m s}^{-1}$, $a = -9.8\ \text{m s}^{-2}$.
- Top: $v = 0$
- $v^2 = u^2 + 2as \Rightarrow s = \dfrac{14.7^2}{19.6} = 11.0\ \text{m}$
- Time up: $0 = 14.7 - 9.8t \Rightarrow t = 1.5\ \text{s}$
- Total (symmetry): $2 \times 1.5 = 3.0\ \text{s}$
Max height $\approx 11.0\ \text{m}$; total time $= 3.0\ \text{s}$.
Where the marks go
- 1 mark: Selects $v^2 = u^2 + 2as$ with $v = 0$ and correct signs
- 1 mark: Correct maximum height $\approx 11.0\ \text{m}$
- 1 mark: Uses $v = u + at$ to find time to the top ($1.5\ \text{s}$)
- 1 mark: Correct total time of flight $3.0\ \text{s}$
Key idea
Choose a positive direction, set $v = 0$ at the peak, and use the symmetry of constant-acceleration motion to halve the work.
Example 2 — Relative velocity of a river crossing
Question
A boat heads directly across a river, pointing its bow straight at the opposite bank with a speed of $3.0\ \text{m s}^{-1}$ relative to the water. The river flows parallel to the banks at $4.0\ \text{m s}^{-1}$. Find the boat's speed relative to the ground and the angle its resultant path makes with the straight-across direction.
Solution
The boat's velocity relative to the ground is the vector sum of its velocity relative to the water and the water's velocity relative to the ground. These two are perpendicular.
Speed: $v = \sqrt{3.0^2 + 4.0^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0\ \text{m s}^{-1}$.
Angle from the across-direction: $\tan\theta = \dfrac{4.0}{3.0}$, so $\theta = \tan^{-1}(1.333) = 53^\circ$ (2 s.f.).
Resultant: $5.0\ \text{m s}^{-1}$ at $53^\circ$ downstream of straight across. Recognise the 3–4–5 triangle and don't add the speeds arithmetically.
The key idea is that velocities add as vectors, not as plain numbers. The boat pushes straight across at $3.0\ \text{m s}^{-1}$ while the current carries it sideways at $4.0\ \text{m s}^{-1}$, and these two directions are at right angles.
Because they're perpendicular, we use Pythagoras for the resultant speed:
$v = \sqrt{(3.0)^2 + (4.0)^2} = \sqrt{25} = 5.0\ \text{m s}^{-1}$.
For the direction, the current is the "opposite" side and the across-velocity is the "adjacent" side, so $\tan\theta = \dfrac{4.0}{3.0} = 1.333$, giving $\theta = 53^\circ$.
So the boat actually travels at $5.0\ \text{m s}^{-1}$, angled $53^\circ$ downstream from where it's pointing. That downstream drift is why you can't reach the point directly opposite if you aim straight across.
Perpendicular velocities; add as vectors.
- $v = \sqrt{3.0^2 + 4.0^2} = \sqrt{25} = 5.0\ \text{m s}^{-1}$
- $\tan\theta = \dfrac{4.0}{3.0} \Rightarrow \theta = 53^\circ$
$5.0\ \text{m s}^{-1}$ at $53^\circ$ downstream of straight across.
Where the marks go
- 1 mark: Recognises the two velocities are perpendicular and adds them as vectors
- 1 mark: Correct resultant speed $5.0\ \text{m s}^{-1}$
- 1 mark: Correct angle $53^\circ$ relative to the across direction
Key idea
Velocities combine as vectors; perpendicular components give a resultant via $v = \sqrt{v_x^2 + v_y^2}$ and a direction from $\tan\theta$.