derivation of kinematic equations yay! lots of math! “math is the best!” – previous physics...
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Derivation of Kinematic Equations
Yay! Lots of math!
“Math is the best!” – Previous Physics Student
“I love word problems!” – Former CHS student
Constant velocity
Average velocity equals the slope of a position vs time graph when an object travels at constant velocity.
avg
xv
t
Displacement when object moves with constant velocity
The displacement is the area under a velocity vs time graph
v
t
x
x v t
Uniform acceleration
This is the equation of the line of the velocity vs time graph when an object is undergoing uniform acceleration.
The slope is the acceleration
The intercept is the initial velocity
f iv at v v
t
v0
v
t
Displacement when object accelerates from rest
Displacement is still the area under the velocity vs time graph. However, velocity is constantly changing.
v
t
This looks like the graph of something that’s speeding up!
You’re right, buddy! Let’s go save the princess!
Displacement when object accelerates from rest
Displacement is still the area under the velocity vs time graph. Use the formula for the area of a triangle.
x 12vt
v
t
v
12area base height
Displacement when object accelerates from rest
From slope of v-t graph
Rearrange to get
Now, substitute for ∆v
in the equation from the last slide
a v
t
v a t
x 12a t t
v
t
v
Displacement when object accelerates from rest
Simplify
Assuming uniform acceleration and a starting time = 0, the equation can be written:
v
t
v
x 12a (t)2
x 12
at 2
Displacement when object accelerates with initial velocityBreak the area up into two parts:
the rectangle representingdisplacement due to initial velocity
v
t
v0
x v 0t
Displacement when object accelerates with initial velocityBreak the area up into two parts:
and the triangle representingdisplacement due to acceleration
x 12a(t)2
v
t
v0
v
Displacement when object accelerates with initial velocitySum the two areas:
Or, if starting time = 0, the equation can be written:
x v 0t 12
a(t )2
x v0t 12at 2
v
t
v0
v
Time-independent relationship between ∆x, v and a
Sometimes you are asked to find the final velocity or displacement when the length of time is not given.
To derive this equation, we must start with the definition of average velocity:
v x
t
Relationship between ∆x, v and a
Another way to express average velocity is:
avg
xv
t
v v f v 0
2
That average is average.
Time-independent relationship between ∆x, v and a
We have defined acceleration as:
This can be rearranged to:
and then expanded to yield :
va
t
vt
a
f iv vt
a
Time-independent relationship between ∆x, v and a
Now, take the equation for displacement
and make substitutions for average velocity and ∆t
x v t
Relationship between ∆x, v and a
x v t
2f iv v
v
f iv vt
a
Relationship between ∆x, v and a
x v t
x vf v0
2vf v0
a
Relationship between ∆x, v and a
Simplify
2f i f iv v v v
xa
2 2
2f iv v
xa
Time-independent relationship between ∆x, v and a
Rearrange
2 2
2f iv v
xa
2 22 f ia x v v
Time-independent relationship between ∆x, v and a
Rearrange again to obtain the more common form:
2 22 f ia x v v
2 2 2f iv v a x
Which equation do I use?
• First, decide what model is appropriate– Is the object moving at constant velocity? Unit 1
– Or, is it accelerating? Unit 2
• Next, decide whether it’s easier to use math or a graph.
• If you use math, follow the table on the board.
Constant velocity
If you are looking for the velocity,– use algebra
– or find the slope of the graph (actually the same thing)
xv
t
Constant velocity
• If you are looking for the displacement,– use algebra
– or find the area under the curve
x v t
v
t
x
Uniform acceleration
• If you want to find the final velocity,– use algebra
• If you are looking for the acceleration– rearrange the equation above
– which is the same as finding the slope of a velocity-time graph
f iv at v
a v
t
Uniform acceleration
If you want to find the displacement,– use the algebraic form
– eliminate initial velocity if the object starts from rest
– Or, find the area under the curve
x v0t 12at 2
v
t
v0
v
If you don’t know the time…
You can solve for ∆t using one of the earlier equations, and then solve for the desired quantity, or
You can use the equation
– rearranging it to suit your needs
vf2 v0
2 2a x
All the equations in one place
constant velocity uniform acceleration
2 2 2f iv v a x
212ix v t at
a v
t
f iv at v
xv
t
x v t