# Particle moving

• Feb 8th 2010, 01:32 PM
poitier
Particle moving
A particle moves along the x-axis so that its velocity at any time t > or equal to zero is given by v(t) = pi(t)-cos((pi)(t)).
Find the acceleration at any time t. Find the minimum acceleration of the particle over the interval [0, 2]. Find the minimum velocity of the particle over the interval [0, 2].

• Feb 8th 2010, 01:49 PM
driegert
So first off. Velocity is the change in distance over time, and acceleration is the change in velocity over time. That is, Velocity is the derivative of the distance function, and acceleration is the derivative of the velocity function. To find the acceleration at any time t we first take the derivative of v(t).

$\frac {dv}{dt}=\pi + \pi sin(\pi t)$

In order to find the minimum velocity we must find the critical points of this derivative (where $\frac {dv}{dt}=0$ and also check the endpoints.

This derivative will be 0 when $sin(\pi t)=-1$ that is when t = 3/2.
So let's check t=0,3/2,2 in our original equation (endpoints and critical values).

$v(0)=-1; v(3/2)=0; v(2)=2\pi -1$

Our minimum velocity would therefore occur when t=0 (i.e. minimum velocity = -1).

Now take the derivative of the acceleration, find critical values, and plug back into the acceleration formula to find minimum accelerations :).

Hope that helps! (and hopefully I didn't mess up any of the algebra)
• Feb 8th 2010, 01:51 PM
icemanfan
For the first part, acceleration is the derivative of velocity. Since we have that velocity $v(t) =\pi t - \cos (\pi t)$,

acceleration $a(t) = v'(t) = \pi + \pi \sin (\pi t)$.