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NEET PHYSICSEasy

Two cars P and Q start from a point at the same time in a straight line and their positions are represented by xP(t)=at+bt2x_P(t) = at + bt^2 and xQ(t)=ft−t2x_Q(t) = ft - t^2. At what time do the cars have the same velocity?

A

(a - f) / (1 + b)

B

(a + f) / 2(b - 1)

C

(a + f) / 2(1 + b)

D

(f - a) / 2(1 + b)

Step-by-Step Solution

Velocity is defined as the rate of change of position with respect to time (v=dxdtv = \frac{dx}{dt}) .

  1. Find the velocity of car P: xP(t)=at+bt2x_P(t) = at + bt^2 vP=dxPdt=ddt(at+bt2)=a+2btv_P = \frac{dx_P}{dt} = \frac{d}{dt}(at + bt^2) = a + 2bt

  2. Find the velocity of car Q: xQ(t)=ft−t2x_Q(t) = ft - t^2 vQ=dxQdt=ddt(ft−t2)=f−2tv_Q = \frac{dx_Q}{dt} = \frac{d}{dt}(ft - t^2) = f - 2t

  3. Equate velocities: Given that the cars have the same velocity: vP=vQv_P = v_Q a+2bt=f−2ta + 2bt = f - 2t

  4. Solve for time (t): Rearrange terms containing tt to one side: 2bt+2t=f−a2bt + 2t = f - a 2t(b+1)=f−a2t(b + 1) = f - a t=f−a2(1+b)t = \frac{f - a}{2(1 + b)}

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