Posted by **Vanessa** on Friday, January 4, 2013 at 10:50am.

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].

- calculus -
**Steve**, Friday, January 4, 2013 at 11:06am
a(t) = dv/dt, so

a(t) = (2pi-5) - 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 pi-5

max accel is at t=1,3 and is 3pi-5

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