Question 71:
moderate
A car accelerates from rest at constant rate of \(2\text{ m/s}^2\) for some time after which it decelerates at a constant rate of \(3\text{ m/s}^2\) to come to rest. If total time taken for the motion is \(40\) seconds then maximum velocity achieved by the car during motion is
Using the relation \(v_{max} = \frac{\alpha \beta}{\alpha + \beta} t\), where \(\alpha = 2\) and \(\beta = 3\) are acceleration and deceleration rates respectively. Substituting the values: \(v_{max} = \frac{2 \times 3}{2 + 3} \times 40 = 48\text{ m/s}\).