# Question #f8288

Apr 27, 2016

$3.0 \times {10}^{4} {\text{ miles" " hour}}^{-} 2$

#### Explanation:

Let us use the expression between velocity, time $t$ and acceleration $a$, assuming a constant force is applied during the interval.

$v = u + a t$
where $v$ is final velocity and $u$ is initial velocity of car. Substituting given values

$60 = 0 + a \times \frac{7.2}{3600}$, seconds to be converted to hours. Solving for $a$
$a = 60 \times \frac{3600}{7.2} = 30000$
$= 3.0 \times {10}^{4} {\text{ miles" " hour}}^{-} 2$