Question

a particle moves along the x-axis so that its velocity at any time t is equal to or > 0 is given by v(t)= (2pie-5)t-sin(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]

Answer 1

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