Question

Initially car A is $ 10.5 $ m ahead of car B. Both start moving at time $ t = 0 $ in the same direction along a straight line. The velocity time graph of two cars is shown in figure. The time when the car B catches the car A will be:
                                                                                                                                                                                                                                                                                                                           

(A) $ t = 21 $
(B) $ t = \dfrac{2}{5} $
(C) $ t = 20 $
(D) None of the above.

Answer 1

Hint :Here, study the given diagram. Use the kinematic equations to find out distance of car B, also use $ a=tan45^0 $ as this is the way to find acceleration. If needed use basic formulas of velocity, distance and time.

Complete Step By Step Answer:
velocity of car A, $ {v_A} = 10.5m{s^{ - 1}} $
Let us now find out the distance travelled by Car A in $ t $ sec, $ {s_A} $ be the distance travelled by car A.
Therefore, $ {s_A} = 10t $
Distance travelled by Car B, distance be $ {s_B} $
 $ {s_B} = \dfrac{1}{2}{t^2} $ (Using formula of kinematic equation, $ s = ut + \dfrac{1}{2}a{t^2} $ and $ a = \tan {45^0} $ )
Now, as per the given condition that the Car A is $ 10.5 $ m ahead of Car B, we have
 $ {s_A} + 10.5 = \dfrac{1}{2}{t^2} $
 $ \Rightarrow 10.5 + 10t = \dfrac{1}{2}{t^2} $
 $ \Rightarrow {t^2} - 20t - 21 = 0 $
 $ \Rightarrow t = \dfrac{{20 \pm \sqrt {{{(20)}^2} - 4( - 21)} }}{2} $ (using, $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ )
 $ \Rightarrow t = 21 $ sec
Thus the time taken by car B to catch car A is $ 21 $ seconds.
Option A is the correct answer.

Note :
As we know that velocity of any body is the result of distance travelled by the boy with respect to time. Here, car A is running forward of car B to catch car A, it has to cover the difference between them in seconds. Finding out distance travelled by car B is not necessary as we have to use a kinematical equation to find time taken.