Get Answers to all your Questions

header-bg qa

A car is standing 200 m behind a bus, which is also at rest.  The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2.  The car will catch up with the bus after a time of :


  • Option 1)

    \sqrt{110}  s

  • Option 2)

    \sqrt{120}  s

  • Option 3)

    10 \sqrt{2}  s

  • Option 4)

    15 s


Answers (1)


As we learnt in

Case of Relative velocity. -

When A and B are moving along a straight line in the same direction.

\underset{V_A}{\rightarrow} = Velocity of object A.

\underset{V_B}{\rightarrow} = Velocity of object B.

Relative velocity  of A wr to B is


\vec{V}_{AB},\vec{V}_{A},\vec{V}_{B} All are in same direction.


- wherein

a)   \vec{V}_{AB}=\vec{V}_{A}-\vec{V}_{B}

        = (20 - 5) ms-1

        = 15 ms-1

b)    \vec{V}_{BA}=\vec{V}_{B}-\vec{V}_{A}

         = (5 - 20) ms-1

         = - 15 ms-1


 Initial separation between bus and car S_{CB}=200 \:meter

a_{CB}=a_C - a_B= 2ms^{-2}

u_{CB} =0 , \:t=?

S_{CB}=u_{CB}+\frac{1}{2}a_{CB}t^2=200=\frac{1}{2} \times 2t^2

t^2=200 \Rightarrow t=10\sqrt{2} sec

Option 1)

\sqrt{110}  s


Option 2)

\sqrt{120}  s


Option 3)

10 \sqrt{2}  s


Option 4)

15 s


Posted by


View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE