Two buses P AN Q start form a point at the same time and move in a straight line and their positions are represented by Xp (t) = at + bt2 and XQ (t) = ft – t2. At what time, both the buses have same velocity?

X P ( t ) = α t + β t 2 V P ( t ) = α + 2 β t - (i) X Q ( t ) = f t − t 2 V Q ( t ) = f − 2 t - (ii) (i) = (ii) α + 2 β t = f − 2 t ⇒ t ( 2 β + 2 ) = f − α t = f − α 2 ( β + 1 )

Two buses P AN Q start form a point at the same time and move in a straight line and their positions are represented by Xp (t) = at + bt2 and XQ (t) = ft – t2. At what time, both the buses have same velocity?

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Physics Class 11th Physics Motion in Straight Line

Two buses P AN Q start form a point at the same time and move in a straight line and their positions are represented by X_p (t) = at + bt² and X_Q (t) = ft – t². At what time, both the buses have same velocity?

Option 1 -

$\frac{α - f}{1 + β}$

Option 2 -

$\frac{α + f}{2 (β - f)}$

Option 3 -

$\frac{α + f}{2 (1 + β)}$

Option 4 -

$\frac{f - α}{2 (1 + β)}$

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

a month ago

Correct Option - 4

Detailed Solution:

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ - (i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ - (ii)
(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$

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Two buses P AN Q start form a point at the same time and move in a straight line and their positions are represented by X_p (t) = at + bt² and X_Q (t) = ft – t². At what time, both the buses have same velocity?

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