Question
Three elephants $$A,B$$ and $$C$$ are moving along a straight line with constant speed in same direction as shown in figure. Speed of $$A$$ is $$5\,m/s$$ and speed of $$C$$ is $$10\,m/s.$$ Initially separation between $$A$$ and $$B$$ is $$'d'$$ and between $$B$$ and $$C$$ is also $$d.$$ When $$'B'$$ catches $$'C'$$ separation between $$A$$ and $$C$$ becomes $$3d.$$ Then the speed of $$B$$ will be
Three elephants $$A,B$$ and $$C$$ are moving along a straight line with constant speed in same direction as shown in figure. Speed of $$A$$ is $$5\,m/s$$ and speed of $$C$$ is $$10\,m/s.$$ Initially separation between $$A$$ and $$B$$ is $$'d'$$ and between $$B$$ and $$C$$ is also $$d.$$ When $$'B'$$ catches $$'C'$$ separation between $$A$$ and $$C$$ becomes $$3d.$$ Then the speed of $$B$$ will be
A.
$$7.5\,m/s$$
B.
$$15\,m/s$$
C.
$$20\,m/s$$
D.
$$5\,m/s$$
Answer :
$$15\,m/s$$
Solution :
$$B$$ catches $$C$$ in time $$t$$ then $$t = \frac{d}{{u - 10}}.$$
Separation by this time has increased by $$'d'$$ between $$A$$ and $$C$$ hence
$$\left( {10 - 5} \right) \times \frac{d}{{\left( {u - 10} \right)}} = d,u = 15\,m/s$$
$$B$$ catches $$C$$ in time $$t$$ then $$t = \frac{d}{{u - 10}}.$$
Separation by this time has increased by $$'d'$$ between $$A$$ and $$C$$ hence
$$\left( {10 - 5} \right) \times \frac{d}{{\left( {u - 10} \right)}} = d,u = 15\,m/s$$
