A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find

(a) Given, the bust starts from rest. Therefore, initial velocity (u) = 0 m/s

Acceleration (a) \( =0.1\,m.s^{-2}\)

Time = 2 minutes = 120 s

Acceleration is given by the equation \( a=\frac{v-u}{t}\)

Therefore, terminal velocity (v) = (at) + u

\( =(0.1\,m.s^{-2}\times 120s)\) \( +0\,m.s^{-1}\)

= 12m.s^-1 + 0 m.s^-1

Therefore, terminal velocity (v) = 12m/s

(b) As per the third motion equation, 2as = \( v^2-u^2\)

Since a = 0.1 m.s^-2 , v = 12 m.s^-1 , u = 0 m.s^-1 , and t = 120s, the following value for s (distance) can be obtained.

Distance, \( s=\frac{v^2-u^2}{2a}\)

\( =\frac{12^2-0^2}{2(0.1)}\)

Therefore, s = 720m.

The speed acquired is 12m.s^-1 and the total distance traveled is 720m