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Question

Two cars P and Q starts 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?

A
fa2(1+b)
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B
af1+b
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C
a+f2(b1)
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D

a+f2(1+b)
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Solution

The correct option is A fa2(1+b)

Position of car P,
xp(t)=at+bt2
Thus, velocity of car P
vp=d[xp(t)]dt=a+2bt

Position of car Q
xQ(t)=ftt2
Thus, velocity of car Q
vQ=d[xQ(t)]dt=f2t

According to the question,
vpt=to= vQt=to

a+2bto=f2to
to=fa2(1+b)


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