First slide

Position Path Length (Distance) And Displacement

Question

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

Moderate

A

$(\frac{α^{2} + β^{2}}{αβ}) t$
B

$(\frac{α^{2} - β^{2}}{αβ}) t$
C

$(\frac{α + β}{αβ}) t$
D

$(\frac{αβ}{α + β}) t$

Solution

Let the car accelerates for a time f, and travels a distance s₁ . Suppose the maximum velocity attained by the car be v. Then
$s_{1} = \frac{1}{2} {at}^{2} and v = {αt}_{2}, t_{1} = v / α$
$∴ s_{1} = \frac{1}{2} \times α \times \frac{V^{2}}{α^{2}} = \frac{V^{2}}{2 α} . . . . (i)$
Let the car decelerates for a time t₂ and travels a distances s₂ .Then
$s_{2} = {vt}_{2} - \frac{1}{2} {βt}_{2}^{2} and 0 = v - {βt}_{2}$
or $t_{2} = \frac{V}{β}$
$∴ s_{2} = v \times (\frac{v}{β}) - \frac{1}{2} β (\frac{v^{2}}{β^{2}})$
or $s_{2} = \frac{v^{2}}{β} - \frac{v^{2}}{2 β} = \frac{v^{2}}{2 β} . . . . . (ii)$
Now, $t_{1} + t_{2} = t or \frac{V}{α} + \frac{V}{β} = t$
$∴ V = \frac{1}{(\frac{1}{α} + \frac{1}{β})} = (\frac{αβ}{α + β}) t$
and S = S₁ + S₂
$= \frac{V^{2}}{2 α} + \frac{V^{2}}{2 β} = \frac{V^{2}}{2} (\frac{1}{α} + \frac{1}{β})$

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