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: Use the kinematic - equation $v^{2}=v_{0}^{2}+2ax$ to find acceleration $a$.

We know $v = 3\ m/s$, $v_{0}=15\ m/s$ and $x = 30\ m$. Substitute into the formula: $3^{2}=15^{2}+2a\times30$. $9 = 225+60a$. $60a=9 - 225=-216$. $a=\frac{-216}{60}=- 3.6\ m/s^{2}$.

Step2: Use the kinematic - equation $v = v_{0}+at$ to find time $t$.

We know $v = 3\ m/s$, $v_{0}=15\ m/s$ and $a=-3.6\ m/s^{2}$. Rearrange the formula for $t$: $t=\frac{v - v_{0}}{a}$. $t=\frac{3 - 15}{-3.6}=\frac{-12}{-3.6}=\frac{10}{3}\approx3.33\ s$.

Answer:

$\frac{10}{3}\ s$