Question

Two cars $P$ and $Q$ start from a point at the same time in a straight line and their positions are represented by $x _{ P }( t )= at + bt ^{2}$ and $x _{ Q }( t )=f t - t ^{2}$. At what time do the cars have the same velocity ?

• A. $\frac{a - f}{1 + b}$
• B. $\frac{a + f}{2( b - 1)}$
• C. $\frac{a + f}{2( 1 + b)}$
• D. $\frac{f - a }{2( 1 + b)}$

Answer 1

The Correct Option is D

We are given two velocity equations for particles P and Q:

\(v_p = \frac{dx_p}{dt} = a + 2bt\)

\(v_Q = \frac{dx_Q}{dt} = f - 2t\)

The condition is that both particles move with the same velocity at some time \(t\), so:

\(v_P = v_Q \Rightarrow a + 2bt = f - 2t\)

Now, let's solve for \(t\):

\(a + 2bt = f - 2t\)

Move the terms involving \(t\) to one side:

\(2bt + 2t = f - a\)

Factor out \(t\):

\(t(2b + 2) = f - a\)

Finally, solve for \(t\):

\(t = \frac{f - a}{2(b + 1)}\)

Conclusion:

The time \(t\) at which the velocities of particles P and Q are equal is given by:

\(t = \frac{f - a}{2(b + 1)}\)