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Q.

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

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a

a+f21+b

b

f-a21+b

c

a-f1+b

d

a+f2b-1

answer is B.

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Detailed Solution

Given, xP(t) = at + bt2  and  xQ(t) = ft - t2Hence vP(t) = a+2bt  and  vQ(t) = f -2tAs vp=vQ,  a+2bt = f - 2t. Therefore  t=f-a21+b.
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