A bus starting from rest moves with a uniform acceleration of 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) = (??) + u
\begin{array}{l}
=\left(0.1 \mathrm{m} \cdot \mathrm{s}^{-2} * 120 \mathrm{s}\right)+0 \mathrm{m} \cdot \mathrm{s}^{-1} \\
=12 \mathrm{m} \cdot \mathrm{s}^{-1}+0 \mathrm{m} \cdot \mathrm{s}^{-1}
\end{array}
Therefore, terminal velocity (v) = 12m/s
(b) As per the third motion equation,
2 a s=v^{2}-u^{2}
Since a=0.1 \mathrm{m} \cdot \mathrm{s}^{-2}, \mathrm{v}=12 \mathrm{m} \cdot \mathrm{s}^{-1}, \mathrm{u}=0 \mathrm{m} \cdot \mathrm{s}^{-1} and t = 120s, the following value for s (distance) can be
obtained.
Distance \mathrm{s}=\frac{v^{2}-u^{2}}{2 a}
=\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.
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