Question

the position of a particle is given by s(t)=sin^2(t)+3 for the interval 0 11 years ago

Answer 1

Velocity is the derivative of position.

s'(t) = v(t) = d/dt (sin^2(t)+3)

Answer 2

yes which is v(t)=2sin(t)cos(t). my problem is what is my next step

Answer 3

I think you need to find when it is at rest, or v=0. Then try a value on both sides of that to see which is positive/negative.

Answer 4

yes but unfortunately i'm a little rusty on my trig so if you could detail out 2sin(t)cos(t)=0 that would be very helpful

Answer 5

Alright. You have a double angle identity right there:
2sin(x)cos(x)=sin(2x)

so then it would be
0=sin(2t)
arcsin(0)=2t

arcsin(0)=0, pi, 2pi

0=2t or pi=2t or 2pi=t
0=t or pi/2=t or pi=t
0 and pi are out of range, so it's just t=pi/2
That should be when it's at rest. Test a point on each side. If the result is positive, that side has positive velocity. If it's negative, that side has negative velocity.

Answer 6

thank you! but i believe the answer will be [0,pi/2)

Answer 7

Yes, that would be correct, but I think it should exclude 0 since that's a point at rest (and it's excluded from the range anyway). So (0,pi/2)? Unless I'm mistaken

Answer 8

isn't the hard [ exclusionary?

Answer 9

I don't know for sure. I thought that [0,pi/2) means 0 <= t < pi/2.