A train starting from rest attains a velocity of 72km/h in 5minutes find (1) acceleration is uniform (2) distance travelled by the train for attaining the velocity 5
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Hi.
given initial velocity u = 0
final velocity v = 72 km/h = 20 m/s
time interval t = 5 min = 300 sec
acceleration a = v-u/t = 20-0/300 =1/15 = 0.066 m/sec2
dist. travelled s = ut +1/2at2 (hlf of a * t square)
= 1/2 * 1/15 * 300 * 300 = 3000 m
given initial velocity u = 0
final velocity v = 72 km/h = 20 m/s
time interval t = 5 min = 300 sec
acceleration a = v-u/t = 20-0/300 =1/15 = 0.066 m/sec2
dist. travelled s = ut +1/2at2 (hlf of a * t square)
= 1/2 * 1/15 * 300 * 300 = 3000 m
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