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 ?
- a-f1+b
- a+f2(b-1)
- a+f2(1+b)
- f-a2(1+b)
Answer
(4) f-a2(1+b)
Solution
vP = dxPdt = a + 2bt
vQ = dxQdt = f - 2t
vP = vQ a + 2bt = f - 2t
2t(1 + b) = f - a
t = f-a2(1+b)
Answer: (4)