Two cars P and Q start from a point at the same time in a straight line and their positions are represented by X _ P (t) = at + b t ^ 2 and X _ Q (t) = ft - t ^ 2 . At what time do the cars have the same velocity?

Correct answer: D. (f - a)/(2 (1 + b)) v _ P = d X _ P dt = a + 2bt v _ Q = d X _ Q dt = f - 2t Now for, v _ P = v _ Q ⇒ a + 2bt = f - 2t ⇒ 2t + 2bt = f - a ⇒ t = (f - a)/(2( b + 1 ))

NEET 2016

Exam: NEET
Year: 2016
Subject: Physics
Chapter: Motion in One Dimension

NEET 2016 · Physics · Motion in One Dimension

Two cars P and Q start from a point at the same time in a straight...

Two cars

P

and

Q

start from a point at the same time in a straight line and their positions are represented by

{X}_{P}\left(t\right) = at + b{t}^{2}

and

{X}_{Q}\left(t\right) = ft - {t}^{2}

. At what time do the cars have the same velocity?

Options

A $\dfrac{a - f}{1 + b}$
B\dfrac{a + f}{2 $\left(b - 1\right)}$
C\dfrac{a + f}{2 $\left(1 + b\right)}$
D\dfrac{f - a}{2 $\left(1 + b\right)}$

Correct answer

D. \dfrac{f - a}{2 $\left(1 + b\right)}$

Step-by-step solution

${v}_{P} = \dfrac{d{X}_{P}}{dt} = a + 2bt$

${v}_{Q} = \dfrac{d{X}_{Q}}{dt} = f - 2t$

Now for, ${v}_{P} = {v}_{Q}$

$⇒ a + 2bt = f - 2t$

$⇒ 2t + 2bt = f - a$

$⇒ t = \dfrac{f - a}{2( b + 1 )}$

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