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Motion in A Straight Line question

2016 · Q141
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Motion in A Straight Line question

2016 · Q141

NEETPhysicsMotion in A Straight LineMCQ+4 / −1
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 + bt2) and xQ(t) = (ft −-− t2).

At what time do the cars have the same velocity ?
  1. A
    a−f1+b{{a - f} \over {1 + b}}1+ba−f​
  2. B
    a+f2(b−1){{a + f} \over {2\left( {b - 1} \right)}}2(b−1)a+f​
  3. C
    a+f2(1+b){{a + f} \over {2\left( {1 + b} \right)}}2(1+b)a+f​
  4. D
    f−a2(1+b){{f - a} \over {2\left( {1 + b} \right)}}2(1+b)f−a​
View written solutionFree

Correct answer: D

For car P,

xP(t) = (at + bt2)

vP(t) = dxp(t)dt{{d{x_p}\left( t \right)} \over {dt}}dtdxp​(t)​ = a + 2bt

Similarly for car Q,

xQ(t) = (ft - t2)

vQ(t) = dxQ(t)dt{{d{x_Q}\left( t \right)} \over {dt}}dtdxQ​(t)​ = f - 2t

When they have same velocity then, vP(t) = vQ(t)

∴\therefore∴ a + 2bt = f - 2t

⇒\Rightarrow⇒ 2t(b + 1) = f - a

⇒\Rightarrow⇒ t = f−a2(1+b){{f - a} \over {2\left( {1 + b} \right)}}2(1+b)f−a​

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