Question

Two cars P and Q start with velocities u and $\frac{u}{2},$ respectively at A but Q, 1 second later. They meet at B with velocities 22 m/s and 31 m/s, respectively. If their accelerations are a and 2a, respectively, find AB (in m)

Answer 1

For car P,
$v^2=u^2+2as$
$484=u^2+2as$ (1)
For car Q,
$961=\frac{u^2}{4}+4as$
$3844=u^2+16as$ (2)
From (1) and (2),
14 as = 3360 $\Rightarrow$ as = 240 (3)
Multiplying equ (1) by, 8 we get
$3872=8u^2+16as$ (4)
On comparing (2) and (4), we get
$28=7u^2\Rightarrow u^2=4$
$\therefore u=\pm 2$
For car P,
v = u + at (By first equation of motion)
$\Rightarrow$ 22 = 2 + at
$\Rightarrow$ at = 20
For car Q,
v = u + at
31 = 1 + 2a(t – 1)
$\Rightarrow$ 31 = 1 + 2at – 2a
$\Rightarrow$ 2a = 10
$\therefore$ $a=5m{s}^2$
Put a = 5 $m{s}^2$ in (3), we get
$s=\frac{240}{5}=48m$