Two-Stage SUVAT Problems | CAIE 9702 Paper 2 Notes

Paper 2

Physics9702

Two-Stage SUVAT Problems

Harder questions contain two linked motions. Finish the first stage, then carry only the shared result into the second.

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 turnAdapted from 9702/12/O/N/25 Q6 · 1 mark

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)

Show worked answer

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\ \text{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: $s_B = +3.6 \text{ m},\; a = +9.81,\; t = 1.7 \text{ s},\; u_B = -|u|$ .
Evaluate the acceleration term first.
$\tfrac12at^2=14.2$
$3.6=-1.7|u|+14.2$
$|u| = \dfrac{14.2 - 3.6}{1.7} = 6.2 \text{ m s}^{-1}$ .

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.

Two-Stage SUVAT Problems

Harder questions contain two linked motions. Finish the first stage, then carry only the shared result into the second.

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 turnAdapted from 9702/12/O/N/25 Q6 · 1 mark

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)

Show worked answer

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\ \text{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: $s_B = +3.6 \text{ m},\; a = +9.81,\; t = 1.7 \text{ s},\; u_B = -|u|$ .
Evaluate the acceleration term first.
$\tfrac12at^2=14.2$
$3.6=-1.7|u|+14.2$
$|u| = \dfrac{14.2 - 3.6}{1.7} = 6.2 \text{ m s}^{-1}$ .

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