Answer:
Explanation:
Give it that,
Initial velocity
u = 22m/s
Deceleration a = - 0.15m/s2
Time taken to travel a station long of 210m
Using equation of motion
Let know the final velocity, when it leaves the station
v² = u²+2as
v² = 22²+2×(-0.15)×210
v² = 484—63
v² = 421
v =√421
v = 20.52m/s
Then,
Using equation of motion to find time taken
v = u + at
20.52 = 22 +(-0.15)t
20.52-22 = -0.15t
-0.15t = -1.48
t = -1.48/-0.15
t = 9.88 sec
