Question

2.7 a car slows down from 15 m/s to 3 m/s over a distance of 36 m. how long did this take?

Explanation:

Step1: Find the average velocity

The average velocity $v_{avg}=\frac{v_i + v_f}{2}$, where $v_i = 15$ m/s and $v_f=3$ m/s. So $v_{avg}=\frac{15 + 3}{2}=\frac{18}{2}=9$ m/s.

Step2: Calculate the time

We know that distance $d = v_{avg}\times t$. Given $d = 36$ m and $v_{avg}=9$ m/s. Rearranging for time $t=\frac{d}{v_{avg}}$. So $t=\frac{36}{9}=4$ s.

Answer:

4 s