ap calc
posted by james on .
a particle moves along the xaxis so that its velocity at any time t is equal to or > 0 is given by v(t)= (2pie5)tsin(piet)
a.find the acceleration at any time t.
b.find the minimum acceleration of the particle over the interval [0,3]
c.find the maximum velocity of the particle over the interval [0,2]

I will read that as
v(t) = (2π  5)t  sin(πt)
= 2πt  5t  sin(πt)
a(t) = 2π  5  πcos(πt)
for a min of a , a ' (t) = 0
π^2 sin(πt) = 0
sin(πt) = 0
πt = 0 or πt = π or πt = 2π
t = 0 , t = 1 , t = 2
a(0) = 2π  5  πcos0 = π  5
a(1) = 2π  5  πcosπ = 2π5+π = 3π5
a(2) = 2π  5  πcos2π = 2π  5  π = π  5
a(3) = ... = 3π  5
looks like the min is π5
c) for max velocity, a(t) = 0
2πt  5t  sin(πt) = 0
sin (πt) = 2πt  5t
very messy equation to solve,
I ran it through Wolfram's "magic" webpage and got
t = 0 and t = .670349
so
v(.670349) = 0
v(0) = sin0 = 0
v(2) = (2π5)(2)  sin 2π
 2.566..  0 = 2.566
check my arithmetic