kinematic equations of motion

Kinematics 1DKinematics 1DKinematic EquationsKinematic Equations

Equation How UsedHow altered when objects start from

rest, v0 = 0

How altered when v is constant,

a = 0

Kinematic EquationsKinematic Equations

The kinematic equations can only be used in problems with uniform (constant magnitude) acceleration.

For constant velocity we already have v

x

t

x v0t 1

2at 2

v 2 v02 2ax

v v0 at

x 1

2at 2

v 2ax

v at

x v0t

v v0

v v0

Relates displacement and time

Relates velocity and time

When time in not mentioned

Zero’s, Plusses, and MinusesZero’s, Plusses, and Minuses

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These are often the trick to the problem and hidden in the text.

Zero Positives and Negatives

x0 Start at origin+ Starts right of origin− Starts left of origin

x Ends at origin+ Ends right of origin− Ends left of origin

Δx Returns to starting point+ Final position is to the right of initial position− Final position is to the left of initial position

v0 Initially at rest+ Moving right− Moving left

v Stops+ Moving right− Moving left

a Constant velocity

+ Moving right (+) and speeding up (+)+ Moving left (−) and slowing down (−)− Moving right (+) and slowing down (−)− Moving left (−) and speeding up (+)

t N/A + Always

Example 1Example 1

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An object initially at rest is displaced 150 m reaching a final speed of 30 m/s.

a. Determine the object’s acceleration.

Making a variable list is a huge step in the right direction.

Look for hidden zero quantities, and picture a coordinate axis to assist in determining the correct sign of each variable.

x0 = 0 m Never said where it started: Assume the origin.

x = Never given, but can be figured out from displacement.

Δx = 150 m Given in the text of the problem.

v0 = 0 m/s Hidden in the words, “initially at rest.”

v = 30 m/s Given in the text of the problem.

a = ? The unknown.

t = Never given.

We can eliminate the first two variables since we already have displacement.

Example 1Example 1

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Δx = 150 m

v0 = 0 m/s

v = 30 m/s

a = ?

t =

Deciding on an equation

1. Type of motion?

If you answer constant velocity, use

If you answer acceleration continue.

2. Time mentioned?

If no, use

If yes continue.

3. Time and displacement mentioned?

If yes, use

4. Time and velocity mentioned?

If yes, use

v

x

t

v2 v

02 2ax

x v

0t

1

2at2

v v

0 at

Accelerating, and time is not mentioned.

v2 v

02 2ax

Example 1Example 1

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Δx = 150 m

v0 = 0 m/s

v = 30 m/s

a = ?

t =

Solve showing the following work

1. Write the original equation (or an equivalent).

2. Rearrange to solve for requested variable.

3. Show substitution of given values.

4. Solve boxing answer and including correct units.

v2 v

02 2ax


v2 v

02 2ax

a

v2 v02

2x

a 30 2

0 2

2 150

a 3 m s2


Example 1Example 1

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b. Determine the time of the motion.

Δx = 150 m

v0 = 0 m/s

v = 30 m/s

a =

t =

Acceleration from part (a) can be added to the list.

1. I’m going to choice to easier (non-quadratic) formula.

2. Rearrange to solve for requested variable.

3. Show substitution of given values.

4. Solve boxing answer and including correct units.

Now time is mentioned, and both velocity and displacement are known.

t

v v0

a

t 30 0

3

t 10 s

x v0t 1

2at 2

v v0 at

v v0 at

3 m/s2

Example 2Example 2

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An object moving at 32.0 m/s decelerates to a stop in a distance of 240 m. Determine the objects acceleration.

Δx = 240 m

v0 = 32.0 m/s

v = 0 m/s

a =

t =


v2 v

02 2ax

v2 v

02 2ax

a

v2 v02

2x

a 0 2

32.0 2

2 240

a 2.13 m s2

Acceleration has two signs.

It is a vector, so direction of motion influences its sign. Since no direction was given I assumed the object was moving in the +x direction. It did not reverse. Direction remained positive.

Acceleration is also a rate. If velocity is increasing (speeding up) acceleration is positive, and if decreasing (slowing) it is negative.

Here the object was decelerating in the positive direction. − + a = −a .

Example 3Example 3

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An object moves at a constant 15 cm/s for 1 minute. Determine the objects displacement.

v

x

t

x 9.0 m

Ask the first question.

Type of motion?

Constant velocity

Don’t waste time with the kinematic equations and all those variables.

It’s the simple equation from the summer homework.

However, watch out for conversions !!!!!!!!!!!!!

x vt

x 0.15 60

How would final speed change if x is doubled to 2x ?

Double x in the derived equation and determine the multiplier for v that will maintain the equality.

Doubling x increases the right side by root two. Therefore, the left side must also be multiplied by root two.

No values, but you can still use a variable list. Simply check of any variable mentioned or implied.

An object uniformly accelerates check off aa

from rest vv0 0 = 0= 0

reaching a speed v check off vv

while moving a distance x check of ΔΔxx

Example 4Example 4

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An object uniformly accelerates from rest reaching a speed v while moving a distance x . Determine its speed when it has moved a distance 2x . Answer in terms of v .

Δx =

v0 =

v =

a =

t =

Accelerating. Time not mentioned.

Substitute v0 = 0 , and rearrange in terms of v .

v2 v

02 2ax

0 (rest)0 (rest)

? v 2a 2x

v 2ax

2v v 2ax

Time is keyTime is key

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1st Using the Quadratic Equation 2nd Avoiding the Quadratic Equation

There are two ways to solve for time, each involving two steps.There are two ways to solve for time, each involving two steps.

v

02 2a x x

0 v0 at

220 02v v a x x Time unknown:

v v

02 2a x x

0 Solve for v :

Result: Two velocities (parabolic). Positive velocity is +x motion and the negative velocity is −x motion. Decide the direction of the final velocity and choose the correct velocity.

x v

0t

1

2at2Main equation:

t

v0 v

02 2a x a

Result: Two times

Why? Quadratic equations are parabolic with 2 x points for every y .

Which time is correct? Choose positive nonzero times. However, if there are two positives it can be tricky to pick.

1

2a

t2 v0 t x 0

t v

0 v0 2

41

2a

x

21

2a

Quadratic:

0v v at Substitute into:

Example 5Example 5

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An object is moving in the +x direction at 25 m/s. At the instant it passes through the origin it encounters an acceleration of 3.0 m/s2 directed opposite its motion. The acceleration ceases to act when the object is 6.0 m to the left of the origin. Determine the elapsed time of the motion.

Δx =

v0 =

v =

a =

t =

Begin by identifying key variables and their correct signs.

6.0 m to the left −6.0 m

+x direction at 25 m/s

+25 m/s

3.0 m/s2 directed opposite

−3.0 m/s2

? s

Determinethe elapsed time

x v

0t

1

2at2

6.0 25 t

1

2 3.0 t2

1.5t2 25t 6 0

t

b b2 4ac

2a

t 25 25 2

4 1.5 6 2 1.5

t

25 625 36

3

t 0.24s or 17s

Negative time makes no sense in real life.

Mathematically it is a point on a parabola if time had run backwards.

Ignore the negative, and choose the positive value.

First we’ll try solving this using the quadratic method.

Example 5Example 5

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An object is moving in the +x direction at 25 m/s. At the instant it passes through the origin it encounters an acceleration of 3.0 m/s2 directed opposite its motion. The acceleration ceases to act when the object is 6.0 m to the left of the origin. Determine the elapsed time of the motion.

Δx =

v0 =

v =

a =

t =

6.0 m to the left −6.0 m

+x direction at 25 m/s

+25 m/s

3.0 m/s2 directed opposite

−3.0 m/s2

? s

Determinethe elapsed time

v2 v

02 2ax

v v

02 2ax

The quadratic is technically superior, but this method has major advantages.

1.Same amount of work.

2.Velocity is calculated.

3.Choosing the correct sign on velocity is just a matter of direction.

4.Students mess up the quadratic 99% of the time. It’s harder to get all the signs correct in the quadratic.

v 25 2

2 3.0 6.0

v 25.7 m/s or 25.7 m/s

If the object is to the left of the origin it must be moving left. Choose the negative velocity.

v v

0 at

t

v v0

a

t 25.7 25

3.0

t 17 s

Now let’s try the other two equations, avoiding the quadratic.

Often Multiple Equations Are NeededOften Multiple Equations Are Needed

Occasionally you may choose an equation that seems obvious, Occasionally you may choose an equation that seems obvious, but then realize that you are missing important variables.but then realize that you are missing important variables.

Example:Example: Given velocity data you are asked to solve for time. Given velocity data you are asked to solve for time.

•You chose an equation containing these variables.You chose an equation containing these variables.

•However, you realize that you are missing acceleration.However, you realize that you are missing acceleration.

•Do not give up. There has to be another way.Do not give up. There has to be another way.

•Turn out that you can solve:Turn out that you can solve:

•But, how does this help?But, how does this help?

•It solves for the variable that was missing in your original equation, and It solves for the variable that was missing in your original equation, and now you are able to solve for time using:now you are able to solve for time using:

The path to the final answer is not always obvious, or done in one step.The path to the final answer is not always obvious, or done in one step.

““Solve what you can solve, and keep going. The answer will Solve what you can solve, and keep going. The answer will eventually turn up.”eventually turn up.”

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v v

0 at

v2 v

02 2ax

v v

0 at

Multi-Stage ProblemsMulti-Stage Problems

Some problems are actually several different problems in one.Some problems are actually several different problems in one.

How are these identified?How are these identified?

If acceleration changes, then it is a new and separate problem.If acceleration changes, then it is a new and separate problem.

Each acceleration has it’s own variable list and equation.Each acceleration has it’s own variable list and equation.

Solve each part separately, but keep the following in mind:Solve each part separately, but keep the following in mind:

•The ending velocity for one part is the initial velocity for the part that The ending velocity for one part is the initial velocity for the part that follows.follows.

•If asked for the total displacement, then solve for the displacement in each If asked for the total displacement, then solve for the displacement in each phase separately and add them together.phase separately and add them together.

•If asked for the total time, then solve for the time in each phase separately If asked for the total time, then solve for the time in each phase separately and add them together.and add them together.

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Example 6Example 6

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A car is at rest at a traffic signal. When the light turns green the car accelerates at 2.4 m/s2 for 15 s. Then the driver maintains a constant velocity for 0.80 km. Finally the car slows to a stop in a distance of 180 m.

Δx =

v0 =

v =

a =

t =

Δx =

v0 =

v =

a =

t =

Δx =

v0 =

v =

a =

t =

This problem has three distinct parts, with three different accelerations

A car is at rest at a traffic signal. When the light turns green the car accelerates at 2.4 m/s2 for 15 s.

2.4 m/s2

“ 2.4 m/s2 ”

the car accelerates at 2.4 m/s2 for 15 s. Then the driver maintains a constant velocity for 0.80 km.

0 m/s2

“constant velocity”

What other variables can be discerned from the problem above?

Are there any important facts we should keep in mind?

The final velocity of one phase becomes the initial velocity for the next.

Constant velocity means that initial and final velocities are equal.

0 m/s

15 s

0 m/s

180 m800 m

−? m/s2

“slows to a stop”

maintains a constant velocity for 0.80 km. Finally the car slows to a stop in a distance of 180 m.

−? m/s2

Example 6Example 6

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v v

0 at

v 0 2.4 15 v 36m s

36 m/s

36 m/s

36 m/s

36 m/s

x v

0t

1

2at2

x 0 15 1

22.4 15 2

x 270m

270 m

v

x

t

Constant VelocityEquation

36 800

t

t 22m s

v2 v

02 2ax

0 2

36 2 2a 180

a 3.6m s2

v v

0 at

0 36 3.6 t t 10s

22 s 10 s

−3.6 m/s2

Δx =

v0 =

v =

a =

t =

Δx =

v0 =

v =

a =

t =

Δx =

v0 =

v =

a =

t =

2.4 m/s2 0 m/s2

0 m/s

15 s

0 m/s

180 m800 m

−? m/s2

Example 6Example 6

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36 m/s

36 m/s

36 m/s

36 m/s

270 m

22 s 10 s

−3.6 m/s2

Δx =

v0 =

v =

a =

t =

Δx =

v0 =

v =

a =

t =

Δx =

v0 =

v =

a =

t =

2.4 m/s2 0 m/s2

0 m/s

15 s

0 m/s

180 m800 m

To find the total time sum the times of all three phases.

To find the total displacement sum the displacements of all three phases.

ttotal

t1 t

2 t

3 15 22 10 47 s

x

totalx

1 x

2 x

3 270 800 180 1250 m

Deriving the Kinematic EquationsDeriving the Kinematic Equations

Acceleration complicates motion problems.Acceleration complicates motion problems.

Galileo discerned the most important kinematic equation from an Galileo discerned the most important kinematic equation from an experiment involving rolling a ball down an incline. He understood experiment involving rolling a ball down an incline. He understood the relationships between the variables in the 1500’s. Calculus had the relationships between the variables in the 1500’s. Calculus had not yet been invented and algebraic equations were originally not yet been invented and algebraic equations were originally determined by experimentation.determined by experimentation.

From From the other equations can be derived. the other equations can be derived.

You will not be responsible for these derivations (one of them uses You will not be responsible for these derivations (one of them uses Calculus), but it is interesting to see how they all relate to one Calculus), but it is interesting to see how they all relate to one another and support one another.another and support one another.

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x v

0t

1

2at2

Deriving the Kinematic EquationsDeriving the Kinematic Equations

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x v

0t

1

2at2

x x

0 v

0t

1

2at2

x v

0t

1

2at2

Constant Velocitya = 0

x v

0t

1

20 t2

x v0t

Constant Velocityv = v0

x vt

v

x

t

Velocity is the derivative of position with respect to time. Need to use position,

not displacement.

v

dx

dt

v

d x0 v

0t

1

2at2

dt

v v

0 at

Combine with

v v

0 at

rearranged as

t

v v0

a

x v

0

v v0

a

1

2a

v v0

a

2

v2 v

02 2ax

kinematic equations of motion

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