KinematicsEasy

Question

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 ?

Options

A.

a-f1+b

B.

a+f2(1-b)

C.

a+f2(1+b)

D.

f-a2(1+b)

Solution

Vp = dxpdt= a + 2bt
VQ = dxQdt=f-2t
VP = VQ
⇒a + 2bt = f - 2t
2t + 2bt = f -  a
⇒ t = f-a2(b+1)

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