Posted by **Savannah** on Friday, January 20, 2012 at 5:33pm.

A particle is moving along the x-axis so that at time t its acceleration is

a(t)=ðcos(ðt)

At time t=1/2, the velocity v of the particle is 1/2.

Find the velocity of the particle at any time t. I think I got that right, as sinðt + C.

Now it wants the minimum velocity of the particle. How do I find that?

- Calculus AB -
**Savannah**, Friday, January 20, 2012 at 5:34pm
Ugh the pi symbol didn't work. All of the ð should be pi. so it would be a(t) = picos(pit)

and my answer should be sin(pit) + C

- Calculus AB -
**Reiny**, Friday, January 20, 2012 at 6:31pm
a(t) = π cos(πt)

v(t) = sin (πt) + c . where v(t) is velocity

when t=1/2, v(1/2) = 1/2

1/2 = sin (π/2) + c

1/2 = 1 + c

c = -1/2

v(t) = sin (πt) - 1/2

for a min of v(t) , v'(t) = 0 , that is, a(t) = 0

πcos(πt) = 0

cos(πt) = 0

πt = π/2 or πt = 3π/2

t = 1/2 or t = 3/2

we already know v(1/2) = 1/2

we need

v(3/2) = sin(3π/2) - 1/2

= -1 - 1/2 = -3/2

so the minimum velocity is -3/2

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