Answer to Question #179790 in Physics for Seemoo

Question

The velocity of a car which is decelerating uniformly changes from 30m/s to 15m/s in 3.3s and covers a distance of 75m. After what further distance will the car finally come to rest?

Answer 1

Find deceleration:

$$a=\frac{v_f-v_i}{t}=\frac{15-30}{3.3}=-4.55\text{ m/s}^2.$$

Find distance it will take to stop from 15 m/s to 0 at this deceleration:

$$d=\frac{v_f^2-v_i^2}{2a}=\frac{0^2-15^2}{2(-4.55)}=24.7\text{ m}.$$