A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation

Options

A. Both will be equal
B. First will be half of second
C. First will be 1/4 of second
D. No definite ratio

Answer

B

Explanation

Let a be the retardation of boggy, then distance covered by it be S. If u is the initial velocity of boggy after deteching from train (i.e., uniform speed of train)
`upsilon^(2)=u^(2)+2as rArr 0 = u^(2)-2as rArr s_(b)=(u^(2))/(2a)`
Time taken by boggy to stop
`upsilon=u+at rArr 0=u-at rArr t=(u)/(a)`
In this time t distance travelled by train `=s_(t)=ut=(u^(2))/(a)`
Hence, ratio `(s_(b))/(s_(t))=(1)/(2)`.