A car is standing 200, m behind a bus,which i | vedclass

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(C) Given: Initial velocity of car $u_C = 0$,initial velocity of bus $u_B = 0$. Acceleration of car $a_C = 4\, m/s^2$,acceleration of bus $a_B = 2\, m

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$10\sqrt{2}\, s$

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(C) Given: Initial velocity of car $u_C = 0$,initial velocity of bus $u_B = 0$. Acceleration of car $a_C = 4\, m/s^2$,acceleration of bus $a_B = 2\, m/s^2$. The initial separation between them is $s = 200\, m$.We use the concept of relative motion. The relative acceleration of the car with respect to the bus is:$a_{CB} = a_C - a_B = 4 - 2 = 2\, m/s^2$.The relative initial velocity is $u_{CB} = u_C - u_B = 0 - 0 = 0$.Using the equation of motion $s = ut + \frac{1}{2}at^2$ for relative motion:$200 = 0 \cdot t + \frac{1}{2} \cdot a_{CB} \cdot t^2$$200 = \frac{1}{2} \cdot 2 \cdot t^2$$200 = t^2$$t = \sqrt{200} = 10\sqrt{2}\, s$.Thus,the car will catch up with the bus after $10\sqrt{2}\, s$.

$A$ 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 an acceleration of $2\, m/s^2$ and the car has an acceleration of $4\, m/s^2$. The car will catch up with the bus after a time of:

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