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Two cars P and Q start from a point at t...

Two cars `P` and `Q` start from a point at the same time in a straight line and their position are represented by `x_p(t) = at + bt^2` and `x_Q (t) = ft - t^2`. At what time do the cars have the same velocity ?

A

1) `frac(a-f)(1+b)`

B

2) `frac(a+f)(2(b-1))`

C

3) `frac(a+f)(2(1+b))`

D

4) `frac(f-a)(2(1+b))`

Text Solution

Verified by Experts

The correct Answer is:
D
Position of the car P at any time t, is
`x_(p) (t)=at+bt^(2)`
`v_(p)(t)=(dx_(p)(t))/(dt)=a+2bt`…(i)
similarly , for car Q,
`x_(q)(t) = ft-(t)^(2)`
`Vv_(Q)(t) = (dx_(q)(t))/(dt)=f-2t`….(ii)
therefore v_(p) (t) = Vv_(q)(t)`(Given)
`therefore a+2bt=f-2t` or `2t(b+1)=f-a`
`thereforet= (f-a)/(2(1+b)`
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