# Particle Acceleration

In previous articles I talked about the position of a particle as well as the velocity of a particle.  For a brief overview, position is the particles physical location within space according to an observer.  While velocity is the rate that the particles position changes within a set period of time.  Both position and velocity values are subjective to the person observing them.

Acceleration on the other hand is the rate that the velocity changes in a certain amount of time.  You can find acceleration by using the equation below.

(Eq 1)  $a = \frac{dv}{dt}=\frac{d^2s}{dt^2}$

dv = Change in velocity

dt = Change in time

To find acceleration you will have to take the derivative of velocity or the double derivative of the particles position in respect to time.  This means that if you have a known acceleration, and for simplicity lets say the acceleration is constant, than you can determine the velocity of the particle as well as the position of the particle in respect to time.  You would do this by taking the integral of the acceleration constant to find velocity, or by taking the double integral of the acceleration constant to find position.  Refer to the two equations below.

(Eq 2)  $v=v_o+a_ct$

vo = initial velocity

a c = constant acceleration

(Eq 3)  $s=s_o+v_ot+\frac{1}{2}a_ct^2$

so = initial position

###### Example

There is a particle travelling at a constant acceleration of 5 $\frac{m}{s^2}$. An observer states that the initial velocity is 10 $\frac{m}{s}$. After 5 seconds has past, what velocity would the particle be traveling at and what would its position be?

###### Solution

From the above statement we can say that the acceleration of the particle is at a constant 5$\frac{m}{s^2}$ and the particle is observed to have a 5 $\frac{m}{s}$ initial velocity. Since the observer didn’t state an initial position, we can assume that it is zero.

First, in order to find the velocity of the particle for a change in time due to an acceleration we will need to take the integral of the acceleration function. Due to the fact the acceleration function is a constant we can use equation 2.  This would result in the following.

$v = 10 \frac{m}{s} + 5 \frac{m}{s^2} (5s) = 35 \frac{m}{s}$

Next, to find the position of the particle we would either need to take the integral of the velocity function or the double integral of the acceleration function.  Again since the acceleration function is constant, we can use equation 3.  This would result in the following.

$s = 0 + 10 \frac{m}{s} (5s) + \frac{1}{2} (5 \frac{m}{s^2}) (5s)^2 = 92.5 m$

The above equations do not consider the direction of acceleration on the Cartesian Coordinate system, and instead only consider the magnitude of the acceleration .  Since we live in a 3-dimensional you will have to express acceleration in the x-direction, y-direction, and z-direction.  When doing this it is not always easy to initially see what the magnitude of the acceleration are.  To find the magnitude of acceleration you would use the equation below.

(Eq 4) $a=\sqrt{a_xi^2+a_yj^2+a_zk^2}$

###### Example

There is an acceleration in the x-direction of 2.5 $\frac{m}{s^2}$, an acceleration of 4 $\frac{m}{s^2}$ in the y-direction, and an acceleration of 1 $\frac{m}{s^2}$ in the z-direction.  What is the total acceleration of the particle.

###### Solution

To find the total acceleration of the particle you would use equation 4.  Refer to the solution below.

$a=\sqrt{(2.5i)^2+(4j)^2+(1k)^2}=4.82\frac{m}{s^2}$

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