train started from rest attains a velocity of 108 km per hour in 2
minutes As given the acceleration to be uniform find the value of acceleration and distance covered
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Train starts from rest so the initial velocity(u)=0m/s
Final velocity(v) =108km/hr=108×5/18m/s
=30m/s
Time =2min=120sec
Now acceleration =v-u÷t=30-0/120=0.25m/s^2
using the eq. of motion
s=ut+1/2at2=(0×120) + (1/2×0.25×120^2)= 0+1800m=1800m
Final velocity(v) =108km/hr=108×5/18m/s
=30m/s
Time =2min=120sec
Now acceleration =v-u÷t=30-0/120=0.25m/s^2
using the eq. of motion
s=ut+1/2at2=(0×120) + (1/2×0.25×120^2)= 0+1800m=1800m
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