Question

The position time graphs for two particles A and
B moving along the same straight line are shown
in figure. The time after which (from when A
starts) B caught A.

​

Answer 1

First, let's calculate the velocity of A and B :

• Velocity is given as slope of position-time graph.

$v_{A} = \tan( {37}^{ \circ} ) = \dfrac{3}{4} \: m {s}^{ - 1}$

$v_{B}= \tan( {53}^{ \circ} ) = \dfrac{4}{3} \: m {s}^{ - 1}$

Now, particle A starts from position = 4m , whereas B starts from position = 0 m and after 4 sec.

So, let time after which B meets A be t :

At that time, position of A and B are same:

$\{v_{A} \times( t + 4) \} + 4 = ( v_{B} \times t)$

$\implies \{ \dfrac{3}{4} \times( t + 4) \} + 4 = ( \dfrac{4}{3} \times t)$

$\implies \{ \dfrac{3}{4} t + 3 \} + 4 = ( \dfrac{4}{3} \times t)$

$\implies \dfrac{3}{4} t + 7= \dfrac{4}{3} t$

$\implies (\dfrac{16 - 9}{12}) t = 7$

$\implies (\dfrac{7}{12}) t = 7$

$\implies t =12 \: sec$

So, A and B meet after 12 sec.