Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by $X_{P}(t)=\alpha t+\beta t^{2}$ and $\mathrm{X}_{\mathrm{Q}}(\mathrm{t})=\mathrm{ft}-\mathrm{t}^{2}$. At what time, both the buses have same velocity?

(D) $\frac{f-\alpha}{2(1+\beta)}$

(A) $\frac{\alpha-\mathrm{f}}{1+\beta}$

(B) $\frac{\alpha+\mathrm{f}}{2(\beta-1)}$

(C) $\frac{\alpha+f}{2(1+\beta)}$

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Kinematics

Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by

X_{P} (t) = α t + β t^{2}

and

X_{Q} (t) = ft - t^{2}

. At what time, both the buses have same velocity?

(A) $\frac{α - f}{1 + β}$
(B) $\frac{α + f}{2 ( β - 1 )}$
(C) $\frac{α + f}{2 ( 1 + β )}$
(D) $\frac{f - α}{2 ( 1 + β )}$ ✓ Correct

Answer: (D)

Solution

Velocity is the derivative of position with respect to time. For bus P:

V_{P} (t) = \frac{d X _{P}}{d t} = α + 2 β t

. For bus Q:

V_{Q} (t) = \frac{d X _{Q}}{d t} = f - 2 t

. Setting

V_{P} = V_{Q}

α + 2 β t = f - 2 t

. Solving for t:

2 β t + 2 t = f - α

, so

t = \frac{f - α}{2 ( β + 1 )}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option

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