Graphs of Motion - NEET Physics Questions
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Graphs of Motion

Question 21: moderate

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 total time elapsed is t, then maximum velocity acquired by car will be:

(1994)

1. \(\frac{(\alpha^2 - \beta^2)t}{\alpha\beta}\)
2. \(\frac{(\alpha^2 + \beta^2)t}{\alpha\beta}\)
3. \(\frac{(\alpha + \beta)t}{\alpha\beta}\)
4. \(\frac{\alpha\beta t}{\alpha + \beta}\)
View Answer

Let (v_{max}) be the maximum velocity. Time to accelerate: \(t_1 = \frac{v_{max}}{\alpha}). Time to decelerate: (t_2 = \frac{v_{max}}{\beta}). Total time (t = t_1 + t_2 = \frac{v_{max}}{alpha} + \frac{v_{max}}{beta} = v_{max}\left(\frac{1}{alpha} + \frac{1}{\beta}\right) = v_{max}\left(\frac{\beta + \alpha}{\alpha\beta}\right)). Solving for \(v_{max}): (v_{max} = \frac{\alpha\beta t}{\alpha + \beta}).