A car is standing 200 m 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 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of :

Question

A car is standing 200 m 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 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of : (A) √110 s  (B) √120 s  (C) 10 √2 s (D) 15 s

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/38bb783d48139c8a983b596b0f18f2af-.png" /> aCB = 2 m/sec 2 200=(1/2)×2t2 t=10√2 second

Q. A car is standing $200 m$ 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 $2 m / s^{2}$ and the car has acceleration $4 m / s^{2}$ . The car will catch up with the bus after a time of :

$110 s$

$120 s$

$102 s$

$15 s$

$a_{CB} = 2 m / se c^{2}$
$200 = \frac{1}{2} \times 2 t^{2}$
$t = 102$ second

Q. 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 :

110​s

120​s

102​s

15s