Physics 9702 · Paper 2

Using the SUVAT Equations

Every SUVAT question gives you three of the five letters and asks for a fourth.
The method is always the same: list the letters, then pick the one equation that leaves out the letter you neither know nor want.

The method

Write u, v, a, s, t and fill in the three values the question gives.
Mark the one you want.
Pick the equation that leaves out the fifth letter, substitute, and solve.

Before you start, choose one and give every vector a sign that matches it. If up is positive, then $g = -9.81 \text{ m s}^{-2}$ ; if down is positive, $g = +9.81 \text{ m s}^{-2}$ .

Choose one positive direction and give every vector a sign that matches it. Both choices work; mixing them in one calculation is what loses the marks.

Worked example

Warm-up: a dropped ball

A ball is dropped from rest and falls 5.0 m. Find its speed just before it lands.

A hand releasing a small ball that falls toward the ground. — Dropped from rest, the ball speeds up over the 5.0 m fall under gravity alone.

Take down as positive: $u = 0,\; s = 5.0 \text{ m},\; a = 9.81 \text{ m s}^{-2}$ ; we want $v$ .
The missing letter is $t$ , so use $v^2 = u^2 + 2as$ .
$v^2 = 0 + 2 \times 9.81 \times 5.0 = 98.1$ , so $v = 9.9 \text{ m s}^{-1}$ .

Answer

v = 9.9 \text{ m s}^{-1}

Worked example

A harder step: thrown straight up

A ball is thrown straight up at 20 m s⁻¹. Find the greatest height it reaches.

The fact to notice: at the very top the ball is still for an instant, so $v = 0$ .
Take up as positive: $u = +20$ , $a = -9.81 \text{ m s}^{-2}$ ; we want $s$ . The missing letter is $t$ , so use $v^2 = u^2 + 2as$ :
$0 = 20^2 + 2(-9.81)s$
$s = \dfrac{400}{19.62} = 20.4 \approx 20 \text{ m}$ .

Answer

s \approx 20 \text{ m}

Your turn— tap to reveal the worked answer (9702/12/O/N/25 Q5)

A car has initial velocity $u$ , accelerates at a constant rate $a$ through a displacement $d$ , and reaches velocity $v$ . Write $u$ in terms of the others. (9702/12/O/N/25 Q5)

Answer: start from $v^2 = u^2 + 2ad$ and rearrange: $u = \sqrt{v^2 - 2ad}$ . The exam also asks for SUVAT backwards, so practise rearranging before numbers appear.

Two-step questions

Harder questions use SUVAT twice: the first step finds a value (often $a$ or $t$ ), and the second step reuses it.

Your turn— tap to reveal the worked answer (9702/12/O/N/25 Q6)

A bicycle brakes with constant deceleration, slowing from 8.0 m s⁻¹ to 6.0 m s⁻¹ over 7.0 m. How much further does it travel before stopping? (9702/12/O/N/25 Q6)

First find the deceleration: $6.0^2 = 8.0^2 + 2a(7.0)$ , so $a = \dfrac{36 - 64}{14} = -2.0 \text{ m s}^{-2}$ .
Now reuse it from 6.0 m s⁻¹ to rest: $0 = 6.0^2 + 2(-2.0)s$ , so $s = \dfrac{36}{4.0} = 9.0 \text{ m}$ .

Answer: 9.0 m further.

Worked example

Exam-style: two objects, one landing time

Object A is released from rest at a height of 14 m. Object B is thrown upward with speed $u$ from a height of 3.6 m. Both reach the ground at the same time. Find that time, and the speed $u$ .

Two stands of different heights, each with a ball: one dropped, the other thrown upward. — Both balls take the same time t: A is dropped from rest at 14 m, B is thrown up from 3.6 m.

The time, from object A

For A (down positive): $s = \tfrac{1}{2}at^2 \Rightarrow 14 = \tfrac{1}{2}(9.81)t^2$ .
$t = \sqrt{28/9.81} = 1.7 \text{ s}$ .

Answer

t = 1.7 \text{ s}

The initial speed of B

For B, keep down positive, so B's upward throw is negative: $s_B = +3.6 \text{ m},\; a = +9.81,\; t = 1.7 \text{ s},\; u_B = -|u|$ .
$3.6 = -|u|(1.7) + \tfrac{1}{2}(9.81)(1.7)^2 = -1.7|u| + 14.2$
$|u| = \dfrac{14.2 - 3.6}{1.7} = 6.2 \text{ m s}^{-1}$ .

Answer

u = 6.2 \text{ m s}^{-1} \text{ (upward)}

(9702/21/M/J/25 Q1(b))

Common ways the exam changes a SUVAT question: reverse it (give the answer, ask for $u$ or $a$ ), chain two stages with different accelerations, or link two objects through a shared time, as above.

Common mistake

Mixing sign conventions mid-question. Choose the positive direction once, write it down, and give every vector (u, v, a, s) a sign that matches it. A habit that prevents this error: state “taking down as positive” before the first line of working.

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Using the SUVAT Equations

Every SUVAT question gives you three of the five letters and asks for a fourth.
The method is always the same: list the letters, then pick the one equation that leaves out the letter you neither know nor want.

The method

Write u, v, a, s, t and fill in the three values the question gives.
Mark the one you want.
Pick the equation that leaves out the fifth letter, substitute, and solve.

Before you start, choose one and give every vector a sign that matches it. If up is positive, then $g = -9.81 \text{ m s}^{-2}$ ; if down is positive, $g = +9.81 \text{ m s}^{-2}$ .

Choose one positive direction and give every vector a sign that matches it. Both choices work; mixing them in one calculation is what loses the marks.

Worked example

Warm-up: a dropped ball

A ball is dropped from rest and falls 5.0 m. Find its speed just before it lands.

Take down as positive: $u = 0,\; s = 5.0 \text{ m},\; a = 9.81 \text{ m s}^{-2}$ ; we want $v$ .
The missing letter is $t$ , so use $v^2 = u^2 + 2as$ .
$v^2 = 0 + 2 \times 9.81 \times 5.0 = 98.1$ , so $v = 9.9 \text{ m s}^{-1}$ .

Answer

v = 9.9 \text{ m s}^{-1}

Worked example

A harder step: thrown straight up

A ball is thrown straight up at 20 m s⁻¹. Find the greatest height it reaches.

The fact to notice: at the very top the ball is still for an instant, so $v = 0$ .
Take up as positive: $u = +20$ , $a = -9.81 \text{ m s}^{-2}$ ; we want $s$ . The missing letter is $t$ , so use $v^2 = u^2 + 2as$ :
$0 = 20^2 + 2(-9.81)s$
$s = \dfrac{400}{19.62} = 20.4 \approx 20 \text{ m}$ .

Answer

s \approx 20 \text{ m}

Your turn— tap to reveal the worked answer (9702/12/O/N/25 Q5)

A car has initial velocity $u$ , accelerates at a constant rate $a$ through a displacement $d$ , and reaches velocity $v$ . Write $u$ in terms of the others. (9702/12/O/N/25 Q5)

Answer: start from $v^2 = u^2 + 2ad$ and rearrange: $u = \sqrt{v^2 - 2ad}$ . The exam also asks for SUVAT backwards, so practise rearranging before numbers appear.

Two-step questions

Harder questions use SUVAT twice: the first step finds a value (often $a$ or $t$ ), and the second step reuses it.

Your turn— tap to reveal the worked answer (9702/12/O/N/25 Q6)

A bicycle brakes with constant deceleration, slowing from 8.0 m s⁻¹ to 6.0 m s⁻¹ over 7.0 m. How much further does it travel before stopping? (9702/12/O/N/25 Q6)

First find the deceleration: $6.0^2 = 8.0^2 + 2a(7.0)$ , so $a = \dfrac{36 - 64}{14} = -2.0 \text{ m s}^{-2}$ .
Now reuse it from 6.0 m s⁻¹ to rest: $0 = 6.0^2 + 2(-2.0)s$ , so $s = \dfrac{36}{4.0} = 9.0 \text{ m}$ .

Answer: 9.0 m further.

Worked example

Exam-style: two objects, one landing time

Object A is released from rest at a height of 14 m. Object B is thrown upward with speed $u$ from a height of 3.6 m. Both reach the ground at the same time. Find that time, and the speed $u$ .

The time, from object A

For A (down positive): $s = \tfrac{1}{2}at^2 \Rightarrow 14 = \tfrac{1}{2}(9.81)t^2$ .
$t = \sqrt{28/9.81} = 1.7 \text{ s}$ .

Answer

t = 1.7 \text{ s}

The initial speed of B

For B, keep down positive, so B's upward throw is negative: $s_B = +3.6 \text{ m},\; a = +9.81,\; t = 1.7 \text{ s},\; u_B = -|u|$ .
$3.6 = -|u|(1.7) + \tfrac{1}{2}(9.81)(1.7)^2 = -1.7|u| + 14.2$
$|u| = \dfrac{14.2 - 3.6}{1.7} = 6.2 \text{ m s}^{-1}$ .

Answer

u = 6.2 \text{ m s}^{-1} \text{ (upward)}

(9702/21/M/J/25 Q1(b))

Common ways the exam changes a SUVAT question: reverse it (give the answer, ask for $u$ or $a$ ), chain two stages with different accelerations, or link two objects through a shared time, as above.

Common mistake