View source for Equations of Motion

''Helena Dedic''

==Derivation of Equations of Motion==


Consider a v-t graph for an object moving with constant velocity. <br>


•	Motion with constant velocity implies that the acceleration is equal to zero and therefore the v - t graph is a horizontal line. <br>
 <math>v(t) = v_o</math><br>

•	The displacement is the area under the v - t graph 

[[image:Eq_of_Motion_1.png‎|top]]<br>


We write <math>\Delta x(t) = v_ot</math> <br>
 
•	Motion with constant acceleration implies that the velocity is a linear function of time. Given that the initial velocity is <math>v_0</math> we can write <br>

 <math>v(t) = v_o + at</math><br>

where the acceleration is the slope of the graph and<math> v_o</math> is the intercept. <br>

•	The displacement is the area under the v - t graph 


[[image:Eq_of_Motion_2.png‎|top]]<br>

We write <br>

 <math>\Delta x(t) = v_ot + \frac{1}{2}at^2</math><br> 
•	The displacement in both cases is <math>\Delta x(t) = x(t)-x_o</math> <br>
 
•	There are two independent equations describing motion with constant acceleration and five variables: <math>\Delta x(t)</math>, v(t), <math>v_0</math>, a and t. Consequently, in any problem we can solve for two of those variables and three other must be given.  <br>

•	We can derive another useful equation by eliminating t from the two equations above.  <br>  <br>

We begin with equation <br>

 <math>v(t) = v_o + at</math><br>

We isolate t:<br>

 <math>t = \frac{v(t) - v_o}{a}</math> <br>

Then we substitute for t in the equation for the displacement: <br>


 <math>\Delta x(t) = v_o \left(\frac{v(t) - v_o}{a}\right) + \frac{1}{2}a \left(\frac{v(t) - v_o}{a}\right)^2</math> <br> 


We will now multiply both sides by 2a. This step will eliminate fractions from this equation. <br>


 <math>2a\Delta x(t) = 2av_o \left(\frac{v(t) - v_o}{a}\right) + 2a\frac{1}{2}a \left(\frac{v(t) - v_o}{a}\right)^2</math> <br> 

This leads to: <br>

 <math>2a\Delta x(t) = 2v_o (v(t) - v_o) + (v(t) - v_o)^2</math> <br>  <br>

Expanding: <br>

 <math>2a\Delta x(t) = 2v_ov(t) - 2v_o^2 + v^2(t) + v_o^2 - 2v_ov(t)</math> <br>  <br>

which leads to: <br>

 <math>2a\Delta x(t) = - v_o^2 + v^2(t)</math> <br>  <br>

This can finally be written in the form: <br> <br>

 <math> v^2(t) =  v_o^2 + 2a\Delta x(t)</math> <br>

==Exercises==

*[[Problem solving using equations of motion]]

==Free Fall==
*[[Free Fall]]

==Projectile Motion==
*[[Projectile Motion]]