Question

Find the position versus time of a particle moving with the given acceleration data. a(t)=3cos(t)-2sin(t)
s(0)=0, v(0)=4

Answer 1

seems to require integration

Answer 2

v=antiderivative of a use v(0)=4 to find integration constant
s=antiderivative of v use s(0)=0 to find integration constant

Answer 3

do you know how to find the antiderivative of 3cos(t)-2sin(t) ?

Answer 4

No not exactly

Answer 5

do you know derivative of sin(t) is cos(t)
so the antiderivative of cos(t) is sin(t)

do you know the derivative cos(t) is -sin(t)
so the derivative of -cos(t) is sin(t)
so the antiderivative of sin(t) is -cos(t)

Answer 6

Yes I am just confused with the number part of it

Answer 7

antiderivative of 3*cos(t) is 3*sin(t)
antiderivative of-2*sin(t) is -2*(-cos(t))

Answer 8

just bring the constant down

Answer 9

then take the antiderivative of the non-constant part

Answer 10

Okay that makes more sense now. So when I am done with that what is the next step?

Answer 11

oops type-os

Answer 12

And then after that I solve for D right?

Answer 13

\[a(t)=3\cos(t)-2\sin(t) \\ v(t)=\text{ antiderivative of } a(t) \\ v(t)=3\sin(t)-2(-\cos(t))+C \\ v(t)=3 \sin(t)+2 \cos(t)+C \\ \text{ we are given } v(0)=4 \\ v(0)=3\sin(0)+2\cos(0)+C \\ 4=3 \sin(0)+2\cos(0)+C \\ \text{ solve for C } \]

Answer 14

yes once we find this v(t)
we need to do the antiderivative of v(t) to find s(t)

Answer 15

then use s(0)=0 to find your D

Answer 16

Okay this makes so much more sense now thank you

Answer 17

np