Module 1 Lesson 5

Module 1—Motion

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In previous lessons you used graphs to assist in analyzing the motion of objects. In simple cases, this is all you need. You’ve seen that graphical analysis is a powerful tool. When used to solve complex motion problems involving acceleration, velocity, and displacement, graphing often takes too long and is prone to error. Using equations can help with these problems, so you will now investigate some equations that are useful in problem solving.

Equations Involving Acceleration, Velocity, and Displacement

To begin, recall the basic equations that describe average velocity and acceleration.

Using only these two equations and some graphical analysis, several other useful equations have been derived for solving problems involving acceleration, velocity, and displacement. You will learn more about these equations and how they are used in the Watch and Listen and as they are used in the sample solutions provided.

Watch and Listen

There are six equations for motion. You can view their derivation or how they are solved in this animation.

Here are the six kinematic equations for motion:

Equation 4

Equation 1

Equation 5

Equation 2

Equation 6

Equation 3

Note that equation 1, , may be used for both uniform and non-uniform motion.

Equations 2 to 6 are only used when the object is undergoing uniformly accelerated motion such as when it is subjected to constant acceleration.

In the equations above,

is initial velocity.
is the displacement.
Δt is the time interval.

is average velocity.
is the final velocity.
is the acceleration.

Now that you have identified these six equations and when they can be used, how do you approach problem solving using these equations?

A Problem-Solving Method

Use the GRASP method (acronym for Given, Required, Analysis, Solution, Paraphrase) that you have seen in the previous lessons in solving kinematic problems with equations:

Given: Read the question. Identify and write down the known variables. Make a diagram, if needed, to keep everything straight.
Required: From the question, extract and write down what is being asked for (the unknown variable).
Analysis: Decide how to best solve for the required quantity. Select and write down the equation that contains the variables listed. (It does not matter what order the variables are in or which one is unknown.) Specify the positive direction, if needed.
Solution: Manipulate the equation, if necessary, so that the unknown is isolated.
Substitute in the known variables along with their units.
Solve for the answer.
Remember to include units and direction, where appropriate (i.e., vector quantities). Round the final answer to the correct number of significant digits, consistent with the rules on page 877 of the textbook.
Paraphrase: Answer the question in a proper English sentence.

The following series of examples demonstrates this method. Reference these examples as you work on the Try This questions for your Module 1: Lesson 5 Assignment.

A photograph shows a quad moving at high speed.

A quad is in motion.

Example Problem 1: A quad travelling at +10.0 m/s accelerates until the final velocity becomes +20.0 m/s. If the time interval was 5.0 s, determine the quad’s displacement.

Given (Known Variables)
Required (Unknown Variable)	the displacement ()
Analysis	The equation that uses all the variables is .
Solution
Paraphrase	The displacement of the quad is +75 m.

Self-Check

SC 1. Go to page 53 of your textbook and complete question 6 of “1.5 Check and Reflect.”

Check your work.

Self-Check Answer

SC 1.

Given

Required

the skidding distance ()

Analysis and Solution

Distance is a scalar quantity, so direction is not a factor. The magnitude of displacement equals distance. Therefore, the equation that uses all the variables is

Paraphrase

The skidding distance is 19.6 m.

A photograph shows an elk standing on grass. There is a mountain in the background.

Read

Read “Example 1.12” on page 47 of your textbook to see an example of manipulating the equation to isolate the unknown variable. If you feel comfortable manipulating variables, cover the solution to “Example 1.12” and calculate the answer. Then check your work.

Self-Check

SC 2. Solve practice problem 2 for “Example 1.12” on page 47 of your textbook.

Check your work.

Self-Check Answer

SC 2.

Given

= 20 km/h [N]
= 1.5 m/s2 [N]

Δt = 9.3 s

Required

the maximum velocity of the elk ()

Analysis

Choose north to be the positive direction. Convert the initial velocity to m/s. The final velocity will have to be converted back to km/h.

Use the equation , and manipulate it to isolate .

Solution

        = (5.56 m/s) + (1.5 m/s2)(9.3 s)
        = 5.56 m/s + 13.95 m/s
        = 19.5 m/s

Paraphrase

The maximum velocity of the elk is 70 km/h [N].

Module 1: Lesson 5 Assignment

Remember to submit the answer to TR 1 to your teacher as part of your Module 1: Lesson 5 Assignment.

The photograph shows a white dove flying.

Try This

TR 1. A homing pigeon starts from rest and accelerates uniformly at +4.00 m/s2 for 10.0 s. What is the velocity after the acceleration?

Example Problem 2: A car starts from rest and accelerates at +5.0 m/s2 until it has travelled +250 m. What is the time interval?

Given (Known Variables)
Required (Unknown Variable)	the time interval (Δt)
Analysis	The equation that uses all the variables is . It will have to be manipulated to isolate Δt. Since = 0.0 m/s, the equation becomes .
Solution	Now use magnitudes only because you cannot divide vectors.
Paraphrase	The time interval is 10 s.

A photograph shows a motorcycle driving along a street. The bottom, front end of the motorcycle can be seen, along with the pant leg and boot of a man riding the motorcycle.

A motorcycle is on the road.

Read

Read pages 48 and 49 of your textbook. Note how the GRASP method is used with the equation in “Example 1.13.”

Self-Check

SC 3. Solve practice problem 2 for “Example 1.14” on page 50 of your textbook.

Check your work.

Self-Check Answer

SC 3.

Given

= 100 km/h

= –0.80 m/s2

Δt = 1.0 min = 60 s

Required

the distance the motorcycle travelled (Δd)

Analysis and Solution

The equation that uses the variables is . Distance is not a vector quantity, so the final answer will not require direction. Convert the initial velocity to m/s.

Paraphrase

The distance the motorcycle travelled is 2.3 x 102 m.

Module 1: Lesson 5 Assignment

Remember to submit the answer to TR 2 to your teacher as part of your Module 1: Lesson 5 Assignment.

Try This

TR 2. A car has an initial velocity of +30.0 m/s and undergoes an acceleration of –5.00 m/s2 for 5.00 s.

Find the displacement after the acceleration.
Find the velocity after the acceleration.

Read

See how another kinematic equation was derived and how it is used to find the answer to a problem by reading the bottom of page 51 and “Example 1.16” on page 52 of your textbook. Cover the solution to “Example 1.16,” and calculate the answer. Then check your work.

Example Problem 3: An object accelerates from +5.00 m/s to +100 m/s. If the object travelled +250 m during this time, what was the acceleration?

Given (Known Variables)
Required (Unknown Variable)	the acceleration ()
Analysis and Solution	The equation containing all the variables is . Manipulate the equation to isolate a. When dividing using vectors, use the scalar form.
Paraphrase	The acceleration was +20.0 m/s2.

Example Problem 4: A ball rolls up an inclined plane with an initial upward velocity of +9.00 m/s and stops rolling upward after 3.00 s. Then it begins to roll back down the plane. What was the displacement after 3.00 s?

Given (Known Variables)
Required (Unknown Variable)	the displacement ()
Analysis and Solution	Choose up the ramp as the positive direction. The equation to use is
Paraphrase	The displacement after 3.00 s was +13.5 m.