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  • I have a doubt, kindly clarify. 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 c

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)

best_answer

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}

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

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

- 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

Incorrect

Option 2)

\sqrt{120}  s

Incorrect

Option 3)

10 \sqrt{2}  s

Correct

Option 4)

15 s

Incorrect

Posted by

prateek

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