A car is standing 200 m behind a bus- which is also rest- The two start moving the same instant but with different forward accelerations- The bus has acceleration 2 m-s-2 and the car has acceleration 4 m-s-2- The car will catch up with the bus after a time of - 10sqrt-2-ssqrt-110-ssqrt-120-s15s

Question

Toppr Admin · Accepted Answer

Car -xA0-and -xA0-bus -xA0-distance - -xA0-200mcar -xA0-ac-4m-s2bus -xA0-aB-2m-s2ac -xA0-with -xA0-respect -xA0-to -xA0-bus - 2m-s2 forwardSo- S-ut-12at2200-0-12-xD7-2-xD7-t2t2-200t-10-x221A-2s

A car is standing 200m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2m/s2 and the car has acceleration 4m/s2. The car will catch up with the bus after a time of : 10√2s√110s√120s15s

Car and bus distance = 200mcar ac=4m/s2bus aB=2m/s2ac with respect to bus = 2m/s2 forwardSo, S=ut+12at2200=0+12×2×t2t2=200t=10√2s

Car and bus distance $=$ $200 m$

car $a_{c} = 4 m / s^{2}$

bus $a_{B} = 2 m / s^{2}$

$a_{c}$ with respect to bus = $2 m / s^{2}$ forward

So, $S = u t + \frac{1}{2} a t^{2}$

$200 = 0 + \frac{1}{2} \times 2 \times t^{2}$

$t^{2} = 200$

$t = 10 \sqrt{2} s$