Posted by **james ** on Sunday, February 17, 2013 at 8:28pm.

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]

- ap calc -
**Damon**, Sunday, February 17, 2013 at 8:54pm
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

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