JEE Main 2022 · 25 June · evening · Single correct · +4 / −1
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 t² and X_Q(t) = ft - t². At what time, both the buses have same velocity?
- (a) (α - f)/(1 + β)
- (b) (α + f)/2(β - 1)
- (c) (α + f)/2(1 + β )
- (d) (f - α)/2(1 + β )
Answer and solution
Answer (d) (f - α)/2(1 + β )
Solution
Physics sir · 4 steps
Same velocity, not same position: differentiate both position equations, then set them equal.
- Velocity of P. Differentiate X_P = α t + β t² with respect to time to get v_P = α + 2β t.
- Velocity of Q. Differentiate X_Q = ft - t² to get v_Q = f - 2t. Note the minus sign stays with the 2t term.
α + 2β t = f - 2t
- Solve for t. Collect the t terms on the left and the constants on the right: 2β t + 2t = f - α, so 2t(1 + β) = f - α.
- So. t = (f - α)/2(1 + β), which is option (d).
The idea to keep. When a question compares velocities, work with v = dx/dt, not with x itself. Differentiate first, then equate, and watch the signs as terms cross the equals sign.
Answer from the NTA final answer key. Solution by PrepWiser.