14. A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is

14. A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is

(a) \( ({\frac{\alpha^2+\beta^2}{\alpha\beta}}) \) t

(b) \( ({\frac{\alpha^2-\beta^2}{\alpha\beta}}) \) t

(c) \( \frac{(\alpha+\beta)t}{\alpha\beta} \)

(d) \( \frac{\alpha\beta t}{\alpha+\beta} \)

Answer :  (d) \( \frac{\alpha\beta t}{\alpha+\beta} \)

\( \alpha = \frac{v_2 - v_1}{t_2 - t_1} \)

\( = \frac{v_{\max}-0}{t_2-0} \)

\( = \frac{v_{\max}}{t_2} \)

\( \Rightarrow t_2 = \frac{v_{\max}}{\alpha} \)

\( \beta = \frac{v_2 - v_1}{t_2 - t_1} \)

\( = \frac{0 - v_{\max}}{t - t_2} \)

\( = -\frac{v_{\max}}{t - t_2} \)

\(\Rightarrow \beta = -\frac{v_{\max}}{t - \frac{v_{\max}}{\alpha}} \)

\( \beta t - \frac{\beta v_{\max}}{\alpha} = -v_{\max} \)

\( \beta t = \frac{\beta v_{\max}}{\alpha} - v_{\max} \)

\( \beta t = v_{\max}\left(\frac{\beta}{\alpha}-1\right) \)

\( \beta t = v_{\max}\frac{\beta-\alpha}{\alpha} \)

\( v_{\max} = \frac{\alpha \beta t}{\beta-\alpha} \)

If \( \beta \) is taken without negative , then

\( v_{\max} = \frac{\alpha \beta t}{\beta+\alpha} \)