A particle move along X direction with constant acceleration.its velocity after 4s is 18m/s and displacement is 56 m.find out its initial velocity and acceleration
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V = 18m/s, s = 56m, t = 4s
a = v-u/t = 18-u/4
Now
s = ut+1/2at^2
56 = u*4 + 1/2(18-u)/4*4^2
56 = 4u + 2(18-u)
56 = 4u + 36 - 2u
56 = 2u + 36
2u = 20
u = 10m/s
Then
a = 18-10/4 = 8/4 = 2m/s^2
a = v-u/t = 18-u/4
Now
s = ut+1/2at^2
56 = u*4 + 1/2(18-u)/4*4^2
56 = 4u + 2(18-u)
56 = 4u + 36 - 2u
56 = 2u + 36
2u = 20
u = 10m/s
Then
a = 18-10/4 = 8/4 = 2m/s^2
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