Asked by sammi
the position of a body at time t sec is s=t^3 - 6t^2 + 9t m. Find the body's acceleration each time the velocity is zero.
Answers
Answered by
Reiny
velocity is ds/dt = 3t^2 - 12t + 9
we need t when velocity is zero, so
3t^2 - 12y + 9 = 0
t^2 - 4y + 3 = 0
(t-1)(t-3) = 0
t = 1 , t = 3
a = dv/dt = 6t - 12
when t = 1, (v=0), a = 6-12 = -6 m/s^2
when t = 3 , a = 18-12 = 6 m/s^2
we need t when velocity is zero, so
3t^2 - 12y + 9 = 0
t^2 - 4y + 3 = 0
(t-1)(t-3) = 0
t = 1 , t = 3
a = dv/dt = 6t - 12
when t = 1, (v=0), a = 6-12 = -6 m/s^2
when t = 3 , a = 18-12 = 6 m/s^2
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