LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
The KINEMATIC EQUATIONS of MOTION
Algebraic Derivations
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
The KINEMATIC EQUATIONS of MOTION
Objectives:
1. Differentiate between a vector quantity and a scalar quantity and state which quantities used in kinematics are vector quantities and which are scalar quantities.
2. Write from memory the equations used to describe uniformly accelerated motion.
3. Complete a data table using information both given and implied in word problems.
4. Use the completed data table to solve word problems.
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Students should understand the special case of motion with constant acceleration so that they:Write down expressions for velocity and position as functions of time, and identify or sketch graphs of these quantities.
Use the equations of motion:
x = Vt
v = v + vo
2
V = V0 + at X = X0 + V0t + (at2)/2 V2 = V02 ­ 2a(X ­ XO )
to solve problems involving 1D motion with constant acceleration.
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
INTERPRETATION and USE of the KINEMATIC EQUATIONS of MOTION
Summarizing results, we have five equations describing various aspects of uniformly accelerated rectilinear motion; these are frequently referred to as the "kinematic relations" for such motion:
v= vo + at
v = v + vo
2
X = 1/2at2 + V0 t + Xo,
V2 ­ Vo 2 = 2a(X­ X0 ).
(The term "kinematic" as used in this context refers to description of motion only, without reference to effects, which produce or alter the motion. Contrasting terms would be "dynamic" or "kinetic," and would imply reference to ideas such as laws of force, effects of inertia ideas concerning causes of motion which have been deliberately excluded from the initial discussion.)
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
The kinematic relations give a complete description of any uniformly accelerated straightline motion. For example, if the motion reverses direction at some instant, the equations say so directly; the same equations describe the motion on either side of the reversal of direction, and we do not resort to two sets of equations, with one set for one direction of the motion and one for the other.
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
The "KINEMATIC EQUATIONS" for UNIFORMLY ACCELERATED MOTION
ALGEBRAIC RELATIONS AMONG X, V, a, and t
Equations are algebraic statements which create meaning for the symbols x, v and a. These statements are definitions; they are not derived by mathematical manipulation from prior connections or relationships. Once we have expressed these definitions in symbols, however, algebraic manipulation leads to the expression of other relationships among the symbols. 6
LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Motion at Constant Acceleration
a = Δv
Δt
but since "Δ" means change
Δv = v ­ vo and a = v ­ vo
t
Δt = t ­ to ; if we always let to = 0, Δt = t at = v ­ vo
V ­ V
o
= at
v = v + at
o
Solving for "v"
This equation tells us how an object's velocity changes as a function of time.
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Starting from rest, you accelerate at 4.0 m/s2 for 6.0s. What is your final velocity? v = vo + at
v = 0 + 4(6)
v = 24 m/s
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
You have an initial velocity of 5.0 m/s. 2 You then experience an acceleration of ­1.5 m/s
for 4.0s; what is your final velocity? v = vo + at
v = 5 ­1.5(4)
v = ­1m/s
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
You have an initial velocity of ­3.0 m/s. 2 for 9.0s; You then experience an acceleration of 2.5 m/s
what is your final velocity? v = vo + at
v = ­3 + 2.5(9)
v = 19.5m/s
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
How much time does it take to accelerate from an initial velocity of 20 m/s to a final velocity of 100 m/s if your acceleration is 1.5 m/s 2 ?
v = vo + at
t = Δv/a
t = (100­20)/1.5
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
How much time does it take to come to rest if your initial velocity is 5.0 m/s and your acceleration is ­2.0 m/s2?
v = vo + at
0 = 5 + ­2t
t = 2.5s
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
An object accelerates at a rate of 3 m/s2 for 6 s until it reaches a velocity of 20 m/s. What was its initial velocity?
v = vo + at
20 = vo + 3(6)
vo = 2m/s 13
LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
Algebraic Derivation of Δx = September 12, 2014
(v + vo )
t
2
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Motion at Constant Acceleration
If velocity is changing at a constant rate, the average velocity is just the average of the initial and final velocities.
v = v + vo
2
And we learned earlier that Δx
v = t
Δx = v + vo
2
t
Some problems can be solved most easily by using these two equations together.
Δx = (v + vo )
t
2
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
This averaging of velocities for bodies moving at constant acceleration is similar to determining a grade in a course. For example, if you have two test grades in the course, your course grade, the average of the two test grades, is the sum of the test grades divided by 2. If we substitute this value of the average velocity into the equation for the displacement derived before, the equation becomes:
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Starting with a velocity of 12 m/s you accelerate to 48 m/s in 6.0s. What is your average velocity?
vavg = (v+v o )/2
vavg = (48+12)/2
vavg = 30m/s
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Starting with a velocity of 12 m/s you accelerate to 48 m/s in 6.0s. Using your previous answer, average velocity
= 30 m/s how far did you travel in that 6.0s?
vavg = Δx/t
Δx = vavg t
vavg = 30(6)
vavg = 180m/s
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Algebraic Derivation of X = X i + V0t+ (1/2)at2
DISPLACEMENT of an OBJECT wrt TIME (Constant Acceleration)
The average velocity does not tell us anything about the body’s acceleration. We would like to express the displacement of the body in terms of its acceleration during a particular time interval, and in terms of its initial velocity, V0 at the beginning of that time interval.
If the acceleration of the moving body is constant, then the average velocity throughout the entire time interval is:
v + v
v = 2 o
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Motion at Constant Acceleration
We can combine these three equations to derive an equation which will directly tell us the position of an object as a function of time.
Δx
v = t
Δx = v t
v + v
v = 2 o
v = vo + at
2. x ­ x o = ½ (v + v o)t
3. x ­ x o = ½ vt + ½ vot
4. x = x o + ½ vot + ½ vt
5. x = x o + ½ vot + ½ (vo + at)t
6. x = x o + ½ vot + ½ vot + ½ at2
x = xo + v o t + ½at2
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Note that V represents the final value of the velocity at the time t, the end of the time interval. But there already exists an equation for the value of V at the time t, namely V = V0 + at Therefore, substituting this equation into the equation above gives:
This kinematic equation represents the displacement x of the moving body at any instant of time t. In other words, if the original velocity and the constant acceleration of the moving object are known, then we can determine the position of the moving object at any time t. This rather simple equation contains a tremendous amount of information.
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
We need to combine three of our equations together to do that v = vo + at The equation we just derived from the definition of acceleration vavg ≡ − 0 The definition of average velocity vavg = ½ (v + v0) The equation for average velocity in the case of constant acceleration Since we have two equations for average velocity, vavg, they must be equal to each other. vavg = vavg We can then substitute in the two different equations for vavg from above: one on the left side of the equals sign and the other on the right [ − 0 ]= [½ (v + v0)] 22
LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Let’s solve this for x: first multiply both sides by t to get it out of the denominator on the left x ­ x0 = ½ (v + v0)t Then add x0 to both sides to get x by itself x = x0 +½ (v + v0)t Distribute t into the parentheses on the right x = x0 + ½ vt + ½v0t Now substitute in our new equation for v: v = vo + at x = x0 + ½ (vo + at) t + ½v0t Distribute ½t into the parentheses x = x0 + ½ vot + ½at2 + ½v0t Combine the two ½ vot terms x = x0 + vot + ½at2 This is another of the key kinematics equations. It allows us to determine where an object will be as time goes by based on a set of initial conditions. In this case, there are three terms: x0 tells us where the object started; vot tells us how fast it was moving initially and how long it’s been traveling; and ½at2, tells us how much its acceleration has affected the distance it has traveled. The reason that t in the last term is squared is that not only does it’s velocity change more as time goes by, it also has had more time for that change in velocity to affect how far it’s gone. 23
LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
An airplane starts from rest and accelerates at a constant rate of 3.0 m/s2 for 30.0 s before leaving the ground. How far did it move along the runway?
x = xo + vot + ½at2
x = ½at2
x = ½(3)30 2
x = 1350m
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
A Volkswagen Beetle moves at an initial velocity of 12 m/s. It coasts up a hill within a constant acceleration of –1.6 m/s2. How far has it traveled after 6.0 seconds?
x = xo + vot + ½at2
x = v
o
t + ½at2
x = 12(6) + ½(­1.6)62
x = 43.2m
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
A motorcycle starts out from a stop sign and accelerates at a constant rate of 20 m/s2. How long will it take the motorcycle to go 300 m?
x = xo + vo t + ½at2
x = ½at2
300 = ½(20)t2
t = 5.5s
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Motion at Constant Acceleration
Algebraic Derivation v2 = vo2 + 2a(x ­ xo)
VELOCITY of an OBJECT w.r.t. DISPLACEMENT (CONSTANT ACCELERATION)
We can also combine equations to eliminate t:
v2 = vo2 + 2a(x ­ xo)
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
In addition to the velocity and position of a moving object at any instant of time, we sometimes need to know the velocity of the moving object at a particular displacement ∆x. To find the velocity as a function of displacement ∆x, we must eliminate time from our kinematic equations.
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Motion at Constant Acceleration
We can also combine these equations so as to eliminate t:
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
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Solving for Vf2, we obtain the kinematic equation:
v2 = vo2 + 2a(x ­ xo)
which is used to determine the velocity V of the moving body at any displacement ∆x and is independent of the variable time, t.
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
A car accelerates from rest to 30m/s while traveling a distance of 20m; what was its accelerate? v2 = vo2 + 2aΔx
v2 = 2aΔx
302 = 2a(20)
a = 22.5m/s2
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Mixed Kinematics Problems
Motion at Constant Acceleration
We now have all the equations we need to solve constant­acceleration problems.
v = vo + at
x = xo + vot + ½at2
v2 = vo2 + 2a(x ­ xo)
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Starting at the position, x0 = 4 m, you travel at a constant velocity of +2 m/s for 6s. a. Determine your position at the times of 0s; 2s; 5s; and 6s. b. Draw the Position versus Time for your travel during this time. c. Draw the Velocity versus Time graph for your trip. 34
LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Starting at the position, x0 = 4 m, you travel at a constant velocity of +2 m/s for 6s.
a. Determine your position at the times of 0s; 2s; 5s; and 6s.
X1=4m
X2=8m
X3=14m
X4=16m
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Starting at the position, x0 = 4 m, you travel at a constant velocity of +2 m/s for 6s.
b. Draw the Position versus Time for your travel during this time.
X (m)
16
4
6 t (s)
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Starting at the position, x0 = 4 m, you travel at a constant velocity of +2 m/s for 6s.
c. Draw the Velocity versus Time graph for your trip.
v (m/s)
2
t (s)
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Position (m)
The position versus time graph, below, describes the motion of three different cars moving along the x­axis.
a. Describe, in words, the velocity of each of the cars. Make sure you discuss each car’s speed and direction.
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
b. Calculate the velocity of each of the cars.
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Position (m)
c. Draw, on one set of axes, the Velocity versus Time graph for each of the three cars.
v
(m/s)
t (s)
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
Summary
• Kinematics is the description of how objects move with respect to a defined reference frame.
• Displacement is the change in position of an object.
• Average speed is the distance traveled divided by the time it took; average velocity is the displacement divided by the time.
• Instantaneous velocity is the limit as the time becomes infinitesimally short.
• Average acceleration is the change in velocity divided by the time.
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
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Summary (continued)
• Instantaneous acceleration is the limit as the time interval becomes infinitesimally small.
• There are five equations of motion for constant acceleration, each requires a different set of quantities.
v = vo + at
x = xo + vot + ½at2
v2 = vo2 + 2a(x ­ xo)
v + v
v = 2 o
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LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
September 12, 2014
The KINEMATIC EQUATIONS of MOTION
Objectives:
4. Differentiate between a vector quantity and a scalar quantity and state which quantities used in kinematics are vector quantities and which are scalar quantities.
Manipulating the mathematical models that we derived into different useful forms of kinematics equations 6. Write from memory the equations used to describe uniformly accelerated motion.
7. Complete a data table using information both given and implied in word problems.
8. Use the completed data table to solve word problems.
c. Solve problems using the kinematic equations.
oSolving kinematics problems.
The mystery of problem solving—how do I do it?
Writing mathematical models (equations) for the constant acceleration model (with physics variables and units). 43
LECTURE_Algebraic_Derivation_of_Kine_Eqs_of_Motion.notebook
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