Question

A truck is traveling at an initial velocity
of +39m/s and it starts slowing down smoothly for 3.3s. It covers a distance of 45m while slowing down. What is the acceleration?​

Answer 1

Answer:

Approximately $-15.37m/s^2$

Explanation:

The acceleration can be found using the formula:

$x_{f} = x_{i} + v_{i} (t)+\frac{1}{2} (a)(t^{2} )$

The work is as shown:

$45m = 0 + 39 m/s (3.3s)+\frac{1}{2} a(3.3s)^2$

$45m=128.7m + 5.445s^2 a$

$-83.7m=5.445s^2a$

$a = -15.37190083m/s^2$

a is about -15.37$m/s^2$