A train normally travels at a uniform speed of 68 km/h on a long stretch of stra

Question

A train normally travels at a uniform speed of 68 km/h on a long stretch of straight, level track. On a particular day, the train must make a 3.0 min stop at a station along the track.
If the train decelerates at a uniform rate of 1.2 m/s^2 and, after the stop, accelerates at a rate of 0.60 m/s^2. How much time is lost because of stopping at the station?
t=__ s A train normally travels at a uniform speed of 68 km/h on a long stretch of straight, level track. On a particular day, the train must make a 3.0 min stop at a station along the track.
If the train decelerates at a uniform rate of 1.2 m/s^2 and, after the stop, accelerates at a rate of 0.60 m/s^2. How much time is lost because of stopping at the station?
t=__ s A train normally travels at a uniform speed of 68 km/h on a long stretch of straight, level track. On a particular day, the train must make a 3.0 min stop at a station along the track.
If the train decelerates at a uniform rate of 1.2 m/s^2 and, after the stop, accelerates at a rate of 0.60 m/s^2. How much time is lost because of stopping at the station?
t=__ s

Explanation / Answer

given,

speed of train = 68 km/h or 18.8889 m/s

train decelerate at the rate of = 1.2 m/s^2

by first equation of motion

v = u + at

since train is stopped its final peed will be 0

0 = 18.8889 - 1.2 * t1

t1 =15.74 sec

while accelerating its final velocity will be 18.8889 and initial velocity will be 0 m/s

18.8889 = 0 + 0.6 * t2

t2 = 31.48 sec

total time = t1 + t2

total time = 15.74 + 31.48

total time lost because of stopping of train = 47.22 sec

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