Question

Two cars P and Q starts from a point at the same time in a straight line and their positions are represented by $ \;{x_P}\left( t \right) = at + b{t^2} $ and $ {x_Q}\left( t \right) = ft - {t^2} $ . At what time do the cars have the same velocity?
A) $ \dfrac{{f - a}}{{2(1 + b)}} $
B) $ \dfrac{{a - f}}{{1 + b}} $
C) $ \dfrac{{a + f}}{{2(b - 1)}} $
D) $ \dfrac{{a + f}}{{2(1 + b)}} $

Answer 1

Hint The velocity of the cars can be calculated as the first time derivative of the position vector. Differentiate both the position functions and equalize them to find the value of the time where both the cars have the same velocity.

Formula used:
 $\Rightarrow Velocity = \dfrac{{dr}}{{dt}} $ where $ r $ is the position vector of the cars

Complete step by step answer
To find the time where both the cars have the same velocity, we should first find the velocity of both the cars. Then, we can find the velocity of the cars P and Q as
$\Rightarrow {v_P}(t) = a\dfrac{d}{{dt}}(t) + b\dfrac{d}{{dt}}({t^2}) $
$\Rightarrow a + 2bt $
In the above equation, $ a $ and $ b $ are constants and hence can be taken out the derivative. Similarly, the velocity of Q can be calculated as
 $\Rightarrow {v_Q}(t) = \dfrac{d}{{dt}}(ft) - \dfrac{d}{{dt}}(2{t^2}) $
 $\Rightarrow f - 2t $
In the above equation, $ f $ is a constant and hence can be taken out of the derivative. For both of these cars to have the same velocity, they should have $ {v_P} = {v_Q} $ at a given time $ t = {t_0} $ which implies
 $\Rightarrow a + 2b{t_0} = f - 2{t_0} $
Solving for $ {t_0} $ , we get
 $\Rightarrow 2b{t_0} + 2{t_0} = f - a $
 $\therefore {t_0} = \dfrac{{f - a}}{{2(b + 1)}} $
So both the cars P and Q will have the same velocity at the time $ \therefore {t_0} = \dfrac{{f - a}}{{2(b + 1)}} $ which corresponds to option (A).

Note
Here we have been given that both the cars travel in the same direction which implies that they have the same direction. If they were travelling in a different direction, we would have to make sure the velocity vectors have the same magnitude and direction for both the cars to have the same velocity since for two vectors to be the same, they must have the same velocity and direction.