Tuesday

March 3, 2015

March 3, 2015

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

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 11:06ama(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

**Answer this Question**

**Related Questions**

Calculus AB - A particle is moving along the x-axis so that at time t its ...

Calculus - At time t >or= to 0, the position of a particle moving along the x...

math - The acceleration of a particle at a time t moving along the x-axis is ...

Calculus - a particle starts at time t = 0 and moves along the x axis so that ...

calculus - 5. A particle moves along the y – axis with velocity given by v(t)=...

Physics-Help please... - The graph give the acceleration a versus time t for a ...

Calculus - The velocity of a particle moving along the x-axis is given by f(t)=6...

Calculous - A particle moves along the c-axis so that at time t its position is ...

Physics - The figure gives the acceleration a versus time t for a particle ...

Physics - Figure 2-41 gives the acceleration a versus time t for a particle ...