Car and mechanics

To help us imagine the situation a bit better, let's start out by saying that the car is driving north and that the bullet enters the back right corner of the car and comes out the front left corner.

The bullet comes in at an angle of $\theta=\tan^{-1}\bigg(\frac{3}{4}\bigg)$ off of the length of the car. Rearranging gives us $\tan\theta=\frac{3}{4}$ . If we extend the length of the car down 4m, over to the right 3m, and then connect back to the back right corner of the car, we get a 3-4-5 triangle where $\theta$ is angle between the hypotenuse and the extended length as described in the original problem.

Now imagine moving the triangle such that its base of length 3m lines up with the width of the car (also of length 3m). The bullet must follow the hypotenuse of triangle, meaning that the front left corner of the car must meet the top left corner of the triangle. The height of the triangle is twice the length of the car, and since the car is 2m long, the car must travel 2m for the bullet to exit at the top right corner of the car.

$\frac{2m}{10m/s}=\boxed{0.2s}$