Question

An object accelerates from rest to 93 m/s over a distance of 49 m. What acceleration did it experience?

Answer 1

Answer:

$88.3ms {}^{ - 2}$

or 88.3m/s^2

Explanation:

Using suvat where we list everything that we are given

s=49m

u=0m/s

v=93m/s

a=?

t=we are not given this value, so we don't use

using a formula that doesn't involve time:

$v {}^{2} = u {}^{2} + 2as$

rearranging to find acceleration by subtracting u^2 on both sides

$v {}^{2} - u {}^{2} = 2as$

then dividing 2s on both sides

$\frac{v {}^{2} - u {}^{2} }{2s} = a$

$\frac{(93) {}^{2} - (0) {}^{2} }{2 \times(49) } = 88.3ms {}^{ - 2}$

so the acceleration is 88.3ms^-2 (1dp)