Question

A subway train accelerates from rest at one station at a rate of 1.20 m/s² for 1/3 of the distance to the next station, then maintains a constant speed for 1/3 of the trip, then starts slowing down at this same rate it accelerated initially for the final 1/3 of the trip. If the stations are 3100 m apart, find

The time of travel between stations

The maximum speed of the train.

Answer 1

here,

total distance between the stations , d= 3100 m

distance where the train accelrates , d1 =1033.33 m

accelrtion , a = 1.2 m/s^2

let the time of accelration be t1

using seccond equation of motion

s = ut +0.5 at^2

1033.33 = 0 + 0.5 *1.2 *t1^2

t1 = 41.5 s

v1 = u1 + a*t1

v1 = 49.8 m/s

for constant speed v1 = 49.8 m/s and distance travelled s2

time taken , t2 = s2/v1

t2 = 20.75 s

time for deaccleration of train is same as that of accelration

therefore , the time of travel , t = 2 *t1 +t2

the time of travel between the stations is 103.75 s

the maximum speed of train is 49.8 m/s