Answer
The average acceleration is $0.038 ~m/s^2$
Work Step by Step
First, we find:
$v_1 = (15 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 4.2 ~m/s$
$v_2 = (65 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 18 ~m/s$
We can use $v_1$ and $v_2$ to find the average acceleration.
$a = \frac{v_2^2 - v_1^2}{2 ~\Delta x} = \frac{(18 ~m/s)^2
- (4.2 ~m/s)^2}{(2)(4000 ~m)} = 0.038 ~m/s^2$
Therefore, the average acceleration is $0.038 ~m/s^2$
