Question

A truck accelerates uniformly from rest to a speed of 13.5 m/s over a distance of 60 m. Determine the acceleration of the truck.

Answer 1

The kinematic equation of motion we need is v2=u2+2as, where:

v = final velocity (ms−1) =13.5m/s

u = initial velocity (ms−1) = 0m/s

a = acceleration (ms−2)=??

s = distance traveled (m) = 60 m

a=(v^2 - u^2)/2s

a = (13.5)^2 - 0^2/2*(60)

a =182.25- 0/120

a=182.25/120

a= 1.51875

a= 1.5 m/s^-2

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Answer 2

Here
u=0 ,v= 13.5 m/s
s= 60m
then acceleration (a) =??
We know that
${v}^{2} + {u}^{2} = 2 \times a \times s$

then
${13.5}^{2} + {0}^{2} = 2 \times a \times 60$

$182.5 = 120 \times a$

$182.25 \div 120 = a$
a=1.51875