Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 $t$ , then the maximum velocity acquired by the car is:

$\left( \frac{\alpha^2 + \beta^2}{\alpha\beta} \right) t$

$\left( \frac{\alpha^2 - \beta^2}{\alpha\beta} \right) t$

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

$\frac{\alpha\beta t}{\alpha + \beta}$

Step-by-Step Solution

Define Variables: Let the maximum velocity acquired be $v_{max}$ . Let the time taken to accelerate be $t_1$ and the time taken to decelerate be $t_2$ . The total time is $t = t_1 + t_2$ .
Analyze Acceleration Phase: The car starts from rest ( $u=0$ ) and accelerates at $\alpha$ . Using the first equation of motion ( $v = u + at$ ) : $v_{max} = 0 + \alpha t_1 \implies t_1 = \frac{v_{max}}{\alpha}$
Analyze Deceleration Phase: The car starts from $v_{max}$ and decelerates at $\beta$ to rest ( $v=0$ ). $0 = v_{max} - \beta t_2 \implies v_{max} = \beta t_2 \implies t_2 = \frac{v_{max}}{\beta}$
Combine and Solve: Substitute $t_1$ and $t_2$ into the total time equation: $t = t_1 + t_2 = \frac{v_{max}}{\alpha} + \frac{v_{max}}{\beta}$ $t = v_{max} \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) = v_{max} \left( \frac{\alpha + \beta}{\alpha\beta} \right)$ $v_{max} = \frac{\alpha\beta t}{\alpha + \beta}$

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