Calculus AB
posted by Savannah on .
A particle is moving along the xaxis 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?

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 
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