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 :

Question

A car starts from rest and accelerates uniformly at 3 m/s\textsuperscript{2. What will be its velocity after 5 seconds?}

Answer 1

To solve this problem, let's analyze the motion of the car and the bus. Both the car and the bus start from rest, but they have different accelerations.

The problem states:

Initial separation distance between the car and the bus is 200\, m.
Acceleration of the bus, a_{bus} = 2 \, m/s^2.
Acceleration of the car, a_{car} = 4 \, m/s^2.

We can calculate the time it takes for the car to catch up to the bus using the kinematic equation for displacement:

The displacement of the bus is given by:

s_{bus} = \frac{1}{2} a_{bus} t^2

The displacement of the car is given by:

s_{car} = \frac{1}{2} a_{car} t^2

For the car to catch up with the bus, the displacement of the car should equal the displacement of the bus plus the initial separation distance:

\frac{1}{2} a_{car} t^2 = \frac{1}{2} a_{bus} t^2 + 200

This simplifies to:

\frac{1}{2} \cdot 4 t^2 = \frac{1}{2} \cdot 2 t^2 + 200

2 t^2 = t^2 + 200

Simplifying further:

t^2 = 200

Therefore,

t = \sqrt{200} = 10 \sqrt{2}

Thus, the time after which the car will catch up with the bus is 10 \sqrt{2}\, s.

This matches with the correct answer given in the options, which is 10 \sqrt{2} \, s.

Answer 2