1. Set up positions of bus and car

Let the bus be initially at position $x=0$. Then the car is $200\,\text{m}$ behind it, so its initial position is $x=-200\,\text{m}$.

Both start from rest at the same instant.

• Bus acceleration: $a_b = 2\,\text{m/s}^2$
• Car acceleration: $a_c = 4\,\text{m/s}^2$

Using $x = x_0 + ut + \frac{1}{2}at^2,$ and since both start from rest, $u=0$.

So,

• Position of bus after time $t$: $x_b = 0 + \frac{1}{2}(2)t^2 = t^2$

• Position of car after time $t$: $x_c = -200 + \frac{1}{2}(4)t^2 = -200 + 2t^2$

2. Condition for catching up

The car catches the bus when their positions become equal: $x_c = x_b$

So, $-200 + 2t^2 = t^2$

$t^2 = 200$

$t = \sqrt{200} = 10\sqrt{2}\,\text{s}$

(We take the positive root since time cannot be negative.)

3. Match with options

$10\sqrt{2}\,\text{s}$ corresponds to Option C.

4. Verification with stored answer

Stored correct answer: C

Our derived answer: C

So the answer agrees with the stored correct answer.