imported>Patrick: /* Derivation of Equations of Motion */

2011-06-05T20:34:44Z

Derivation of Equations of Motion

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''Helena Dedic''

==Derivation of Equations of Motion==

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

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

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

[[image:Eq_of_Motion_1.png‎|top]] 

We write <math>\Delta x(t) = v_ot</math> 

• 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 

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

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

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

[[image:Eq_of_Motion_2.png‎|top]] 

We write 

<math>\Delta x(t) = v_ot + \frac{1}{2}at^2</math> 
• The displacement in both cases is <math>\Delta x(t) = x(t)-x_o</math> 

• 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. 

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

We begin with equation 

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

We isolate t: 

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

Then we substitute for t in the equation for the displacement: 

<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> 

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

<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> 

This leads to: 

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

Expanding: 

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

which leads to: 

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

This can finally be written in the form: 

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

==Exercises==

*[[Problem solving using equations of motion]]

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

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

Equations of Motion - Revision history

imported>Patrick: /* Derivation of Equations of Motion */