Rajdhani Express starting from rest attains a velocity of 108 km per hour in 20 minutes find the acceleration and the distance travelled by the train for attaining this velocity assuming the acceleration to be uniform
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Time taken = 20 mins = 20/60 hrs = 1/3 hr
From first equation of motion,
v=u+at (v=108,u=0,t=1/3 hr)
a = (v-u)/t = (108-0)/1*3 = 108*3 = 324 km/hr^2
1 km/hr^2 = 5/324 m/s^2
324 km/hr^2 = 324*5/324 = 5 m/s^2
From third equation of motion,
v^2=u^2 + 2aS (v=108,u=0,a=324)
11664=0+2*324*S
S=11664/2/108=18 km
Hope this helps :)
(^ means power)
From first equation of motion,
v=u+at (v=108,u=0,t=1/3 hr)
a = (v-u)/t = (108-0)/1*3 = 108*3 = 324 km/hr^2
1 km/hr^2 = 5/324 m/s^2
324 km/hr^2 = 324*5/324 = 5 m/s^2
From third equation of motion,
v^2=u^2 + 2aS (v=108,u=0,a=324)
11664=0+2*324*S
S=11664/2/108=18 km
Hope this helps :)
(^ means power)
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