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Physics/kinematic equations
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Expert: James J. Kovalcin - 2/2/2008
Question
i wanted to know how the kinematic equations were
derived.
Vf = Vi + at
Vf^2 = Vi^2 + 2ax Vi = initial velocity
x = ViT + .5at^2 a = acceleration
x = displacement
vf = final velocity
t = time
How did they come up with these equations?
Get the answer below
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Answer
In my classes students develop these equations in the
laboratory. What my students do is take a block of wood,
attach it to a timing device [which records the position
of the block as a function of time] and then drop the
block out of the 2nd floor window.
They them make 3 graphs: displacement vs time, velocity
vs time and acceleration vs time. Displacement is
derived directly from the timing device [a "tickertape"
timer], velocity is derived by taking the change in
displacement and dividing by the time interval while
acceleration is derived by taking the change in velocity
and diving by the time interval.
The resulting displacement-time graph is a parabola in
the 1st quadrant opening upward, the velocity-time graph
is a diagonal straight line with a positive slope and
the acceleration-time graph turns out to be a horizontal
straight line. Analysis of these three graphs can then
be used to develop the kinematics equations.
Acceleration-time graph
If you look at the area under the acceleration-time
graph you will notice that it is a rectangle. The area
of a rectangle is equal to the length times the width.
In this case this area will be acceleration x time. The
units of acceleration x time will be m/s^2 x s = m/s.
This implies that the area under the acceleration-time
graph is equal to the velocity because m/s are the units
for velocity.
Velocity-time graph
Since this graph is a straight, diagonal line the
equation describing the line will be of the form y=mx+b
where y is the velocity, x is the time, m is the slope
of the line and b is the y-intercept - in this case the
initial velocity. Therefore the equation of the line is
Vf=a*t+Vo [your first equation]. [The velocity in this
situation will have an initial value if the block of
wood is thrown downward rather than being dropped as was
actually done in the lab.]
If you now look at the area under the velocity-time
graph you will notes that the area is in the shape of a
triangle. The area of a triangle is 1/2*base*height. In
the case of the velocity-time graph this area is
1/2*velocity*time=1/2*v*t. If you multiply velocity by
time the resulting units will be m/s*s=m, which is the
unit for displacement so this becomes d=1/2*v*t. But I
showed above that the area under the acceleration-time
graph was v=a*t [remember it was a rectangle]. If you
substitute this in for the velocity v in the triangular
equation you get d=1/2*[a*t)*t=1/2*a*t^2.
All of this assumes that the block was dropped. If the
block was thrown downward instead the diagonal straight
line would have a y intercept. Now the area under the
velocity-time time graph is made up of a triangle plus a
rectangle. We have already figured out the area of the
triangle d=1/2*a*t^2.
The rectangle will have an area equal to height times
width; in this case velocity times time. The height of
the rectangle is the y intercept, the initial velocity
of the "thrown" block, while the width is the time. If
you multiply these together Vo*t the resulting units
will be m/s*s=m, again, the units for displacement.
Therefore this area is a displacement d=Vo*t.
If you add these two areas together you will get the
total displacement of the thrown block of wood, Df=1/2*a*t^2+Vo*t.
Finally, if the thrown block's initial position was
measured relative to some arbitrary origin you will need
to add an initial displacement Do to find the final
displacement, Df=1/2*a*t^2+Vo*t+Do. [Your last equation]
Finally, your middle equation is derived by combining
the other two equations.
First square both sides of the velocity equation.
Vf=a*t+Vo becomes Vf^2=[a^2*t^2+2*a*t*Vo]+Vo^2
Next, used the displacement equation
Df=1/2*a*t^2+Vo*t [Do is assumed to be zero]
Multiply everything by 2*a
2*a*(Df=1/2*a*t^2+Vo*t) which then becomes
2*a*Df=[a^2*t^2+2*a*t*Vo] notice that this is the same
as what is withing the square brackets above,
substitute!
Vf^2=[a^2*t^2+2*a*t*Vo]+Vo^2 which then becomes
Vf^2=[2*a*Df]+Vo^2 And there is your middle equation.
[By the way, I never use this equation in my classes
since it is not needed (although it may be useful in a
few very narrow cases - my philosophy is to have the
students use the least information necessary to solve
any problem.)]
I hope this isn't too much!
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