A train is moving with a velocity of 90 km h-!. It is
brought to stop by applying the brakes which
produce a retardation of 0.5 m s. Find : (i) the
velocity after 10 s, and (ii) the time taken by the to come to rest
Answers
v = u + at
v = 90×5/18 + (-0.5)×10 = 25 - 5
v = 20 m/s
for train to stop , v =0
0 = 25 + (-0.5)t
0.5t = 25
t = 25/0.5 = 50s
A train is moving with a velocity of 90 km/hr. It is brought to stop by applying the brakes which produce a retardation of 0.5 m/s².
We have to find the velocity and time taken by the train.
To convert km/hr into m/s. Multiply the value by 5/18.
Initial velocity = 90*5/18 = 25 m/s and Retardation is -0.5 m/s².
(i) For velocity:
Using the First Equation Of Motion
v = u + at
Substitute the known values,
→ v = 25 + (-0.5)(10)
→ v = 25 - 5
→ v = 20
Therefore, the final velocity of the train is 20 m/s.
(ii) For time taken:
Using the First Equation Of Motion,
v = u + at
When the train comes to rest then it's final velocity become 0 m/s.
So, substituting the values,
→ 0 = 25 + (-0.5)t
→ -25 = -0.5t
→ 50 = t
Therefore, the time taken by the train is 50 sec.