Given position functions

$X_P(t)=\alpha t+\beta t^2$ $X_Q(t)=ft-t^2$

Find velocities by differentiating w.r.t. time

For bus $P$: $v_P(t)=\frac{dX_P}{dt}=\alpha+2\beta t$

For bus $Q$: $v_Q(t)=\frac{dX_Q}{dt}=f-2t$

Set velocities equal

Since both buses have the same velocity at that instant, $\alpha+2\beta t=f-2t$

Solve for $t$

Bring the $t$ terms together: $2\beta t+2t=f-\alpha$

$2(\beta+1)t=f-\alpha$

Therefore, $t=\frac{f-\alpha}{2(\beta+1)}$

Match with the options

$t=\frac{f-\alpha}{2(1+\beta)}$

This matches Option D.

Comparison with stored correct answer

Stored correct answer: D

Derived answer: D

So, the derived answer agrees with the stored correct answer.