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.
![{\displaystyle v(t)=v_{o}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/920a977bfd973c782d5288a921a7672cb6189749)
• The displacement is the area under the v - t graph
![Eq of Motion 1.png](/gwikis/pwiki/images/9/9c/Eq_of_Motion_1.png)
We write
• Motion with constant acceleration implies that the velocity is a linear function of time. Given that the initial velocity is
we can write
![{\displaystyle v(t)=v_{o}+at}](https://wikimedia.org/api/rest_v1/media/math/render/svg/873446367733f47b7f954d27eca9a61df0eb8f12)
where the acceleration is the slope of the graph and
is the intercept.
• The displacement is the area under the v - t graph
![Eq of Motion 2.png](/gwikis/pwiki/images/6/6e/Eq_of_Motion_2.png)
We write
![{\displaystyle \Delta x(t)=v_{o}t+{\frac {1}{2}}at^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b476799d6b1a7e3471e114009216efe1253003d)
• The displacement in both cases is
• There are two independent equations describing motion with constant acceleration and five variables:
, v(t),
, 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
![{\displaystyle v(t)=v_{o}+at}](https://wikimedia.org/api/rest_v1/media/math/render/svg/873446367733f47b7f954d27eca9a61df0eb8f12)
We isolate t:
Then we substitute for t in the equation for the displacement:
We will now multiply both sides by 2a. This step will eliminate fractions from this equation.
This leads to:
Expanding:
which leads to:
This can finally be written in the form:
Exercises
Free Fall
Projectile Motion