Asked by Anonymous
The displacement of a point is given by s = 9.0t3 + 2.0t2 + 6, where s is in meters and t in seconds. Find the acceleration in m/s2 at the instant when the velocity of the particle is 15.0 m/s.
Answers
Answered by
mathhelper
s = 9.0t3 + 2.0t2 + 6
v = ds/dt = 27t^2 + 4t
a = dv/dt = 54t + 4
so when v = 15
15 = 27t^2 + 4t
27t^2 + 4t - 15 = 0
Using the quad formula, I got t = appr .675 and a negative
sub t = .675 into my acceleration formula and you got it
v = ds/dt = 27t^2 + 4t
a = dv/dt = 54t + 4
so when v = 15
15 = 27t^2 + 4t
27t^2 + 4t - 15 = 0
Using the quad formula, I got t = appr .675 and a negative
sub t = .675 into my acceleration formula and you got it
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