Exercise 4.5
A motorbike moving with initial velocity m and constant acceleration stops after travelling m. Find the acceleration of the motorbike and the time taken to come to a stop.
NCERT’s answer
$\displaystyle 4$ m \(\displaystyle s^{- 2}\) in the direction opposite to the velocity; $\displaystyle 7$ s
Acceleration \(\displaystyle = -4\ \text{m s}^{-2} \) (opposite to the direction of velocity); time taken \(\displaystyle = 7\ \text{s} \).
Given: \(\displaystyle u = 28\ \text{m s}^{-1} \), \(\displaystyle v = 0\ \text{m s}^{-1} \) (it stops), \(\displaystyle s = 98\ \text{m} \).
Using Eq. (4.4c): \(\displaystyle v^{2} = u^{2} + 2as \Rightarrow 0 = (28\ \text{m s}^{-1})^{2} + 2 \times a \times 98\ \text{m} \)
\(\displaystyle a = \frac{-784\ \text{m}^{2}\text{s}^{-2}}{196\ \text{m}} = -4\ \text{m s}^{-2} \)
The minus sign says the acceleration is directed opposite to the velocity — the motorbike is slowing down.
Using Eq. (4.4a): \(\displaystyle 0 = 28\ \text{m s}^{-1} + (-4\ \text{m s}^{-2}) \times t \Rightarrow t = \frac{28}{4}\ \text{s} = 7\ \text{s} \)