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Calculus1 18 Online
OpenStudy (anonymous):

I just wanna get my answer checked:

OpenStudy (anonymous):

The height at time t (s) of a weight, oscillating up and down at the end of a spring is given below. s(t) = 400 + 90sin(t)

OpenStudy (anonymous):

a) Find the velocity at t = π/3 s. b) Find the acceleration at t = π/3 s

OpenStudy (anonymous):

so for a) my equation i had was 90sin(t) and b) 90cos(t)

OpenStudy (anonymous):

wondering if those are right.

hartnn (hartnn):

for velocity, u differentiate once, for acceleration, u differentiate twice.

hartnn (hartnn):

so NO

hartnn (hartnn):

first derivative of s(t) = 400 + 90sin(t) is ??

OpenStudy (anonymous):

90cos(t) ??

hartnn (hartnn):

yes, thats your velocity

hartnn (hartnn):

and acceleration ?

OpenStudy (anonymous):

-90sin(t)

hartnn (hartnn):

thats correct also :)

hartnn (hartnn):

now just put t values

OpenStudy (anonymous):

so the answers are: A) 45 b) -45 sqrt3 ???

hartnn (hartnn):

yes.

OpenStudy (anonymous):

woo hoooo. : )

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