A car accelerates from rest at a constant rate $\alpha$ for some time, after which it decelerates at a…

A car accelerates from rest at a constant rate $\alpha$ for some time, after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $\mathrm{t}$, then the maximum velocity acquired by the car is
  1. $\left(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\right) \mathrm{t}$
  2. $\left(\frac{\alpha^{2}-\beta^{2}}{\alpha \beta}\right)$
  3. $\frac{(\alpha+\beta) t}{\alpha \beta}$
  4. $\frac{\alpha \beta \mathrm{t}}{\alpha+\beta}$

Solution



In fig., $\mathrm{AA}_{1}=\mathrm{v}_{\max }=\alpha \mathrm{t}_{1}=\beta \mathrm{t}_{2}$
But $\mathrm{t}=\mathrm{t}_{1}+\mathrm{t}_{2}=\frac{\mathrm{v}_{\max }}{\alpha}+\frac{\mathrm{v}_{\max }}{\beta}$
$=\mathrm{v}_{\max }\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\mathrm{v}_{\max }\left(\frac{\alpha+\beta}{\alpha \beta}\right)$
or, $v_{\max }=t\left(\frac{\alpha \beta}{\alpha+\beta}\right)$ ,

Asked in: JEE Mains - Motion In One Dimension - Test 3

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