A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance.The ratio of distance covered by the boggy and distance covered by the train when the boggy has stopped is
Answers
Answered by
2
Answer
search in Google and best of luck
Answered by
0
Answer:
1/2
Explanation:
Let a be the retardation of boggy then distance covered by it is S. If u is the initial velocity of boggy after detaching From train
So,v^2=u^2+2as
0= u×u-2as(because retardation is there)
s(boggy) = _u×u_
2a. [equation 1 ]
time taken by boggy to stop
v= u+at
0=u-at
t= u/a
in this time t distance travelled by train
s(train)= ut=u×u/a. [ equation 2 ]
by 1 and 2
s(boggy)/ s(train)=1/2
Similar questions