Question

A delivery van is travelling at a constant speed of 15ms^-1 when the driver sees a pedestrian crossing 50m ahead of him. He takes exactly 1 sec before he apploes the brakes as hard as he can. It takes 3 sec for him to stop.

Will the van be able to stop before he reaches the pedestrian crossing?

Answer 1

Using the kinematic formula, find the distance traveled by the van. In the first second, the motion was with constant speed, hence the distance is:

$$d_1 = 15m/s\cdot 1s = 15m$$

During the next 2s the motion was with constant acceleration. Thus, the distance is:

$$d_2 = \dfrac{v+v_0}{2}t = \dfrac{15m/s\cdot 2s}{2} = 15m$$

where $v =0 m/s$ is the final speed, $v_0 = 15m/s$ is the initial speed, and $t = 2s$.

Thus, the total distance:

$$d = 30m$$

and the van will be able to stop before he reaches the pedestrian crossing.