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 + bt^ 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 v _ P = v _ Q ⇒ a + 2bt = f - 2t 2t + 2bt = f - a ⇒ t = (f - a)/(2( b + 1 ))

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Motion in One Dimension

NTA Abhyas 2020 · 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 + bt^{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

{v}_{P} = {v}_{Q}

⇒ a + 2bt = f - 2t

2t + 2bt = f - a

⇒ t = \dfrac{f - a}{2( b + 1 )}

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