the possession as function of time of particle moving along the x-axis given by X(t)=45+10t+2.5t^2 meter. where X in meter and t in second. A, what is average speed b/n t=0 and t=2second,B,instantaneous velocity at the end of 2sec,C.distance covered during 5sec,D.accelereation acquired by the particle
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i dont know...........
yesufmuhammed1160:
plese every one help me to the answer of thies question
x = a + bt^2
v = 2bt
Velocity at t = 0 is 0
Velocity at t = 2s is = 2*2.5*2 = 10 m/s
At t = 2s
x = 8.5 + 2.5*2^2
x = 18.5m
At t = 4s
x = 8.5 + 2.5*4^2
x = 48.5m
Average velocity = (48.5 - 18.5) / 2 = 18m/s
v = 2bt
Velocity at t = 0 is 0
Velocity at t = 2s is = 2*2.5*2 = 10 m/s
At t = 2s
x = 8.5 + 2.5*2^2
x = 18.5m
At t = 4s
x = 8.5 + 2.5*4^2
x = 48.5m
Average velocity = (48.5 - 18.5) / 2 = 18m/s
this answer and my question are d/t so it is not answer
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