Question

A Jeep starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of    110m. Determine the acceleration of the Jeep.​

Answer 1

d = 110m

t=5.21s

acceleration =

$\frac{d}{t}$

a=

$\frac{110}{5.21}$

a

$= 21.11$

or a ~ 21

acceleration = 21 m/s²

Answer 2

Explanation:

Here,

u=0

v=?

d=110m

t=5.21sec

then ,v=d/t.

=110/5.21

=21.11m/sec

now,acceleretion=v-u/t

=21.11-0/5.21

= 4.05 metre per second square