a body starts from the origin and moves along the X-axis such that the velocity at any instant is given by v=4t^3-2t where t is in second and velocity in m/s. find the acceleration of the particle when it is at a distance of 2m from the origin?
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distance = intigration of velocity to respect t
= ∫(4t^3 - 2t).dt
= (4t^4)/4 - (2t^2)/2
2 = t^4 - t^2
take log to both side
log2 = logt^4 - logt^2
0.3 = 4 log(t) - 2 log(t)
0.3 = 2 log(t)
0.15 = log(t)
if we take antilog
t = 1.41 sec
acceleration = d(v)/dt = 12t^2 - 2 =12(1.41)^2 - 2 = 12 x 2 - 2 = 22 m/s^2
regards
= ∫(4t^3 - 2t).dt
= (4t^4)/4 - (2t^2)/2
2 = t^4 - t^2
take log to both side
log2 = logt^4 - logt^2
0.3 = 4 log(t) - 2 log(t)
0.3 = 2 log(t)
0.15 = log(t)
if we take antilog
t = 1.41 sec
acceleration = d(v)/dt = 12t^2 - 2 =12(1.41)^2 - 2 = 12 x 2 - 2 = 22 m/s^2
regards
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