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 XP(t) = αt + βt2 and XQ(t) = ft – t2. At what time, both the buses have same velocity?
- \(\frac{α-f}{1+β}\)
- \(\frac{α-f}{2(β+1)}\)
- \(\frac{α-f}{2(1+β)}\)
- \(\frac{f-α}{2(1+β)}\)
The Correct Option is A
Solution and Explanation
The correct option is(D): \(\frac{f-α}{2(1+β)}\)
XP = αt + βt2
XQ = ft – t2
∴ VP = α + 2βt
VQ = f – 2t
∵ VP = VQ
⇒ α + 2βt = f – 2t
⇒ \(t=\frac{f-α}{2(1+β)}\).
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