Question

A car accelarates from rest at constant rate of 2ms for some time.Then it retards at a constant rate of 4ms and comes to rest.Calculate the maximum speed attained by the car if it remains in motion for 3 seconds.

Answer 1

Answer on Question #43269, Physics, Other

A car accelerates from rest at constant rate of $2\,\mathrm{ms}^{-2}$ for some time. Then it retards at a constant rate of $4\,\mathrm{ms}^{-2}$ and comes to rest. Calculate the maximum speed attained by the car if it remains in motion for 3 seconds.

Solution:

**Given:**

$$\begin{array}{l} a_1 = 2\,\mathrm{m/s^2}, \\ a_2 = -4\,\mathrm{m/s^2}, \\ t = 3\,\mathrm{s}, \\ v = ?, \end{array}$$

For the first period of motion the acceleration is

$$\begin{array}{l} a_1 = \frac{v - v_0}{t_1} = \frac{v}{t_1} \\ t_1 = \frac{v}{a_1} \end{array}$$

For the second period of motion the acceleration is

$$\begin{array}{l} a_2 = \frac{0 - v}{t_2} = -\frac{v}{t_2} \\ t_2 = \frac{-v}{a_2} \end{array}$$

From given

$$t = t_1 + t_2 = \frac{v}{a_1} - \frac{v}{a_2} = v \left(\frac{1}{a_1} - \frac{1}{a_2}\right)$$

Thus,

$$\begin{array}{l} v = \frac{t}{\left(\frac{1}{a_1} - \frac{1}{a_2}\right)} \\ v = \frac{3}{\frac{1}{2} + \frac{1}{4}} = 4\,\mathrm{m/s} \end{array}$$

**Answer:** $v = 4\,\mathrm{m/s}$