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`

Step-by-step 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+(1)/(2)at^2`

Where

s=displacement

u=intial velocity

a=acceleration

t=time

here
`t=30\ s`

Substituting values we get

`80=0+(1)/(2)* a* (30)^2`

`a=(2* 80)/(900)`

`a=(160)/(900)=0.178\ m/s^2`