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 ?

Options

A. `(f-a)/(2(1+b))`
B. `(a-f)/(1+b)`
C. `(a+f)/(2(b-1))`
D. `(a+f)/(2(1+b))`

Answer

A

Explanation

`v_p = (dx_p)/(dt) = a + 2bt`
and `v_Q = (dx_Q)/(dt) = f - 2t`, given, `v_p= v_Q`
` therefore `a + 2bt = f - 2t rArr t = (f-a)/(2(b+1))`