calculus
posted by Vanessa .
A particle is moving along the x axis so that its velocity at any time t ≥0 is give by v(t) = (2 pi  5)t  sin(pi t)
A. Find the acceleration at ay 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].

a(t) = dv/dt, so
a(t) = (2pi5)  pi cos(pi t)
as usual, max/min involve derivatives, so max a(t) occurs where da/dt = 0
da/dt = pi^2 sin(pi t)
which is zero at t = 0,1,2,3
how do we know which is min or max? 2nd derivative of a(t), which is
pi^3 cos(pi t)
so, since cos(pi t) > 0 at t=3/2,
min accel is at t=0,2 and is pi5
max accel is at t=1,3 and is 3pi5 
a) a(t) = v'(t) = 2πcos(2πt)

ops that's wrong heres right answer
A. a(t) = d/dt (v(t)) = d/dt [(2π5)t  sin(πt)] = 2π  5  πcos(πt)
...
B. To find the minimum a(t), we'll derive it and equal to zero:
d/dt (a(t)) = π²sin(πt) = 0 > sinπt = 0 > πt = 0 + 2k*π, k belongs to the integers
t = 2kπ/π = 2k
Since the v(t) is for any t>0 and the given interval is from 0 to 3, the only possible value for k is 1, therefore t = 2*1 = 2 s.
Minimum a(t) is a(2) = (2π  5)  πcos(2π) = 2π  5  π = π  5
...
C. Max speed: derive v(t) and equal it to zero.
d/dt (v(t)) = a(t) = 0
2π  5  πcos(πt) = 0
πcos(πt) = 2π 5
cos(πt) = 2  (5/π)
πt = acos (2  5/π)
t = (1/π)acos(2  5/π) ~ 1.63 s
v(0.366) = (2π  5)*1.63  sin(π*1.63) ~ 3.01 m/s
DISREGARD MY FIRST POST
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