Question

A car is travelling with a speed of 72 km/h on a straight horizontal road . it is followed by another car B which is moving with velocity of 36km/h. When the distance between them is 25m , the car A is given a deceleration of 2ms‐². After how much time will B catch up with A?

Answer 1

velocity of car A, $v_A$ = 72 km/h = 72 × 5/18 = 20 m/s

velocity of car B, $v_B$ = 36 km/h = 36 × 5/18 = 10 m/s

Distance between Car A and car B , $S_0$ = 25m

deceleration of car A, $a_A$ = -2m/s²

Let after time t, car B will catch up car A.

using relative formula,

$s_{rel}=s_0+u_{rel}t+\frac{1}{2}a_{rel}t^2$

when car B catches up the car B, $S_{rel}=0$

$u_{rel}=v_A-v_B=20-10=10m/s$

$a_{rel}=a_A-a_B=-2-0=-2m/s^2$

so, 0 = 25 + (10)t + 1/2(-2)t²

or, 25 + 10t - t² = 0

or, t² - 10t - 25 = 0

or, t = {10 ± √(100 + 25×4)}/2

or, t = (5 ± 5√2) but t ≠ (5 - 5√2)s because (5 - 5√2) is negative value and time can't be negative.

so, t = (5 + 5√2)s