CAIE A-Level Mathematics AS 4.2.4 Motion under Gravity

[Maximum number: 6]

A car travels along a straight road with constant acceleration $a \mathrm{~ms}^{-2}$ , where a>0. The car passes through points A, B and C in that order. The speed of the car at A is $u \mathrm{~ms}^{-1}$ in the direction A B. The distance B C is twice the distance A B. The car takes 8 seconds to travel from A to B and 10 seconds to travel from B to C.

(a)

Find u in terms of a.

[ 4 ]

Using $s=ut+\frac12at^{2}$ ,
$AB=8u+32a,\quad BC=10u+130a,\quad AC=18u+162a.$
Since BC=2AB, 10u+130a=2(8u+32a), giving u=11a.

(b)

Find the speed of the car at C in terms of a.

[ 2 ]

$v=11 a+a \times 18$

M1

For use of v=u+a t or other complete suvat method Using their u in terms of a, e.g.
$\begin{aligned} v^{2}=(11 a)^{2}+2 a(18 \times 11 a+162 a)\left[=841 a^{2}\right], v=\sqrt{(11 a)^{2}+2 a(18 \times 11 a+162 a)}\left[=\sqrt{841 a^{2}}\right] . \end{aligned}$

Speed =29 a

A1FT

FT their expression for v so their u+18 a.

Note: If answer to part (a) is $u=-\frac{7}{3} a$ , then speed $=\frac{47}{3} a$ .

CAIE A-Level Mathematics AS 4.2.4 Straight Lines Questions

Exam points

CAIE A-Level Mathematics AS 4.2.4 Straight Lines Questions question 1

Question (a)

Question (b)