A train is travelling at a speed of 590 km /h. Brakes are applied so as to to produce a uniform acceleration of - 0.5 m s^2. Find how far the train will go before it is brought to rest.
Answers
Answered by
0
Explanation:
initial velocity (u) = 90 km/hour = 25 m.s-1
Terminal velocity (v) = 0 m.s-1
Acceleration (a) = -0.5 m.s-2
Find out
How far the train will go before it is brought to rest
Formula
As per the third motion equation, v2-u2=2as
Therefore,
distance traveled by the train (s) =(v2-u2)/2a
s = (02-252)/2(-0.5) meters
= 625 meters
The train must travel 625 meters at an acceleration of -0.5 ms-2 before it reaches the rest position.
Answered by
0
Answer:
Hello sis.
Would you like to be my friend.
Can I get your intro...
Similar questions