A particle is moving as given by the data below:
v(t)=3sin(t)-7cos(t); s(0)=0

s(t)=?

I got s(t)=-3cos(t)-7sin(t)-3 but it's wrong. Help?

Question

A particle is moving as given by the data below:
v(t)=3sin(t)-7cos(t); s(0)=0

s(t)=?

I got s(t)=-3cos(t)-7sin(t)-3 but it's wrong. Help?

anonymous · Answer

$\frac{d}{dt}s=v$ implies $s=\int v\,dt$, so $-3\cos 0-7\sin 0+C=-3+C=0$ or $C=3$. So you just got the sign wrong.

whpalmer4 · Answer

v(t) = s'(t)
s(0) = 0

Assuming your indefinite integral was $s(t) = -3\cos(t) - 7\sin(t) + c$, let's put in our initial condition and solve for $c$.

$$s(0) = -3\cos(0) - 7\sin(0) + c = 0$$
$$0 = -3(1) - 7(0) + c$$

anonymous · Answer

Oh gosh. Such a careless error, thank you so much for clearing it up!

whpalmer4 · Answer

Negative signs are the work of the devil, I say :-)