Asked by Branny
The path of a particle moving in a straight line is given by s=t³-6t²+9t+4 find s and a when v=0
Answers
Answered by
Damon
ds/dt = v = 3 t^2 - 12 t + 9
d^2s/dt^2 = 6 t - 12 = a
when v = 0
3 t^2 - 12 t + 9 = 0
t^2 - 4 t + 3 = 0
(t - 3) (t-1) = 0
so
when t = 1 and when t = 3 , v = 0
when t = 1
s = 1 - 6 + 9 + 4 = 8
a = 6 - 12 = -6
when t = 3
s = 27 - 54 + 27 + 4 = 4
a = 18-12 = 6
d^2s/dt^2 = 6 t - 12 = a
when v = 0
3 t^2 - 12 t + 9 = 0
t^2 - 4 t + 3 = 0
(t - 3) (t-1) = 0
so
when t = 1 and when t = 3 , v = 0
when t = 1
s = 1 - 6 + 9 + 4 = 8
a = 6 - 12 = -6
when t = 3
s = 27 - 54 + 27 + 4 = 4
a = 18-12 = 6
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