Question

c)
If from rest, Amirul starts to walk to Boon Chun's house and reach there in 30 seconds, what is
his acceleration?

Answer 1

Answer$:0.178\ m/s^2$

Explanation:

Given

Amirul starts from rest(u=0) to reach Boon chun house which is 80\ m away from School

acceleration of Amirul is given by

$s=ut+\frac{1}{2}at^2$

Where

s=displacement

u=intial velocity

a=acceleration

t=time

here $t=30\ s$

Substituting values we get

$80=0+\frac{1}{2}\times a\times (30)^2$

$a=\frac{2\times 80}{900}$

$a=\frac{160}{900}=0.178\ m/s^2$