(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

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

**(a)** the speed acquired,

**(b) **the distance traveled.

A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

What is the quantity which is measured by the area occupied below the velocity-time graph?

What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?