Answer[tex]:0.178\ m/s^2[/tex]
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
[tex]s=ut+\frac{1}{2}at^2[/tex]
Where
s=displacement
u=intial velocity
a=acceleration
t=time
here [tex]t=30\ s[/tex]
Substituting values we get
[tex]80=0+\frac{1}{2}\times a\times (30)^2[/tex]
[tex]a=\frac{2\times 80}{900}[/tex]
[tex]a=\frac{160}{900}=0.178\ m/s^2[/tex]
