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NCERT Solutions Class 9 Science Solutions for Exercise Motion - Intext Questions-5 in Chapter 8 - Motion

Question 5 Motion - Intext Questions-5

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) = (𝑎𝑡) + 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.

Answer:
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