A car accelerates from 31.0m/s to rest over a displacement of 130.0m. What is the car's acceleration, and how long did it take to stop?

Oct 11, 2017

$a = - 3.70 m {s}^{- 2} \left(2 \mathrm{dp}\right)$

$t = 8.39 s \left(2 \mathrm{dp}\right)$

Explanation:

we have

$u = 31 m {s}^{- 1}$

$v = 0 m {s}^{- 1}$

$s = 130 m$

to find $a \text{ & } t$

$\textcolor{b l u e}{{v}^{2} = {u}^{2} + 2 a s}$

$0 = {31}^{2} + 2 \times a \times 130$

$0 = 961 + 260 a$

$\therefore a = - \frac{961}{260} = - 3.70 m {s}^{- 2} \left(2 \mathrm{dp}\right)$

$\textcolor{b l u e}{v = u + a t}$

$0 = 31 - 3.70 t$

$t = \frac{31}{3.70}$

$t = 8.39 s \left(2 \mathrm{dp}\right)$