Exercise 4.4
A car starts from rest and its velocity reaches m in s. Find the average acceleration and the distance travelled in these s.
NCERT’s answer
$\displaystyle 4$ m \(\displaystyle s^{- 2}\) in the direction of velocity, $\displaystyle 72$ m
Average acceleration \(\displaystyle = 4\ \text{m s}^{-2} \), in the direction of the velocity; distance travelled \(\displaystyle = 72\ \text{m} \).
Given: \(\displaystyle u = 0\ \text{m s}^{-1} \) (starts from rest), \(\displaystyle v = 24\ \text{m s}^{-1} \), \(\displaystyle t = 6\ \text{s} \).
Using Eq. (4.3c): \(\displaystyle a = \frac{v - u}{t} = \frac{24\ \text{m s}^{-1} - 0\ \text{m s}^{-1}}{6\ \text{s}} = 4\ \text{m s}^{-2} \)
The magnitude of velocity is increasing, so the acceleration acts along the direction of motion.
Using Eq. (4.4b): \(\displaystyle s = ut + \frac{1}{2}at^{2} = 0 + \frac{1}{2} \times 4\ \text{m s}^{-2} \times (6\ \text{s})^{2} = 72\ \text{m} \)