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]

v = (2 pi  5) t  sin (pi t)
a)
a = dv/dt = (2 pi 5)  pi cos (pi t)
b)
when t = 0, pi cos (pi t) is pi and that is as low as a gets
or more formally
da/dt = 0 = pi sin (pi t)
sin (pi t) = 0 when t = 0 , 1 , 3
then
min a = (2 pi 5)  pi = pi  5
c)
v is max or min when a = 0
(2 pi 5)  pi cos (pi t) = 0
cos (pi t) = 2  5/pi = .4084
pi t = 1.15 radians
t = .366
but pi t can also be 1.15 radians which is + 5.13 radians
t = 1.63 than will give me the bigger v
v = (2 pi  5) t  sin (pi t)
v = (2 pi 5)(1.63)  sin 1.63 pi
= 3