To determine the time at which both buses [tex]\(P\)[/tex] and [tex]\(Q\)[/tex] have the same velocity, we first need to find the velocities of both buses as functions of time [tex]\(t\)[/tex].
1. Velocity of Bus [tex]\(P\)[/tex]:
- The position of bus [tex]\(P\)[/tex] is given by [tex]\( X_p(t) = \alpha t + \beta t^2 \)[/tex].
- To find the velocity, we need to take the derivative of [tex]\(X_p(t)\)[/tex] with respect to time [tex]\(t\)[/tex]:
[tex]\[
V_p(t) = \frac{dX_p(t)}{dt} = \frac{d}{dt} (\alpha t + \beta t^2) = \alpha + 2\beta t
\][/tex]
2. Velocity of Bus [tex]\(Q\)[/tex]:
- The position of bus [tex]\(Q\)[/tex] is given by [tex]\( X_Q(t) = f t - t^2 \)[/tex].
- Similarly, we take the derivative of [tex]\(X_Q(t)\)[/tex] with respect to time [tex]\(t\)[/tex]:
[tex]\[
V_Q(t) = \frac{dX_Q(t)}{dt} = \frac{d}{dt} (f t - t^2) = f - 2t
\][/tex]
3. Equate Velocities:
- To find the time [tex]\(t\)[/tex] at which both buses have the same velocity, we set [tex]\(V_p(t)\)[/tex] equal to [tex]\(V_Q(t)\)[/tex]:
[tex]\[
\alpha + 2\beta t = f - 2t
\][/tex]
- Rearrange this equation to solve for [tex]\(t\)[/tex]:
[tex]\[
\alpha + 2\beta t + 2t = f
\][/tex]
[tex]\[
\alpha + t(2\beta + 2) = f
\][/tex]
[tex]\[
t(2\beta + 2) = f - \alpha
\][/tex]
[tex]\[
t = \frac{f - \alpha}{2\beta + 2}
\][/tex]
- Simplify the denominator:
[tex]\[
t = \frac{f - \alpha}{2(1 + \beta)}
\][/tex]
Therefore, the time at which both buses [tex]\(P\)[/tex] and [tex]\(Q\)[/tex] have the same velocity is:
[tex]\[
t = \frac{f - \alpha}{2(1 + \beta)}
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{\frac{f - \alpha}{2(1 + \beta)}}
\][/tex]
So the correct option is:
(d) [tex]\(\frac{f - \alpha}{2(1 + \beta)}\)[/tex].
