a boy with an initial velocity of 30ms-1 acceleration uniformly at rate of 10ms-1 until attains a velocity 60ms-1. what is the distance covered during the period?
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Answered by
2
given things are : u(initial velocity)=30ms⁻¹ ,v(final velocity)=60ms⁻¹ and a(acceleration)=10ms⁻²
now v²-u²=2ah where h is the distance
60²-30²=2×10×h
3600-900=20h
2700=20h
2700/20=h
270/2=h
135m=h
now v²-u²=2ah where h is the distance
60²-30²=2×10×h
3600-900=20h
2700=20h
2700/20=h
270/2=h
135m=h
Answered by
1
V^2-U^2=2as
60^2-30^2 = 2*10*s
2700/20=s
s= 135m
60^2-30^2 = 2*10*s
2700/20=s
s= 135m
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