Module 1—Motion

Module 1: Lesson 5 Assignment

Remember to submit the answers to LAB 1, LAB 2, LAB 3, LAB 4, and LAB 5 to your teacher as part of your Module 1: Lesson 5 Assignment.

Lesson 5 Lab: Could You Be a Goalie for the NHL?

Complete “1-8 QuickLab: Could You Be a Goalie for the NHL?” on page 57 of your textbook. In step two of the procedure, have the zero end of the ruler just above your hand. In step 6, record where the top of your hand is on the ruler.

LAB 1. Complete steps 1–7 of “Procedure” on page 57 of your textbook. For step 6, you should complete at least three trials. List your trial results.

LAB 2. Answer question 1 on page 57 of your textbook.

LAB 3. Answer question 2 on page 57 of your textbook.

LAB 4. Answer question 3 on page 57 of your textbook.

LAB 5. Answer question 4 on page 57 of your textbook.

Example Problem 5: The driver of a car travelling east at 30.0 m/s applies the brakes to generate an acceleration of 4.00 m/s2 west. If the final speed of the car was 5.00 m/s east, how far did the car travel in the acceleration period?

Given
Required	the distance travelled (Δd)
Analysis and Solution	Choose east as the positive direction. The equation that contains all the variables is . Manipulate the equation to isolate Δd. When dividing using vectors, use the scalar form.
Paraphrase	The distance travelled was 109 m.

Self-Check

SC 4. Solve a problem like the investigators of the Air France crash had to grapple with by answering practice problem 1 of “Example 1.16” on page 52 of your textbook.

Module 1: Lesson 5 Assignment

Remember to submit the answer to TR 3 to your teacher as part of your Module 1: Lesson 5 Assignment.

Try This

TR 3. A rocket sled that was initially at rest reaches a final speed of +30.0 m/s over a displacement of +45.0 m.

1. Find the acceleration.
2. Find the time it took to travel the first 45.0 m.

Read

What happens when the acceleration is caused by gravity? Read from the top of page 56 through the end of “Example 1.17” on the top of page 58.

Example Problem 6: How fast will a falling object travel after 7.00 s if it is dropped from rest?

Given
Required	the final velocity ()
Analysis and Solution	Choose the positive direction to be upwards. The equation, that will have to be manipulated to get , is         = 0.0 + (–9.81 m/s2)(7.00 s)     = –68.7 m/s
Paraphrase	The final velocity of the object will be 68.7 m/s downward.

Example Problem 7: A man standing on the roof of a building throws a stone downward at 20.0 m/s. The stone hits the ground after 5.00 s. How tall is the building?

Given
Required	the height of the building ()
Analysis and Solution	Choose the positive direction to be upwards. The distance the stone falls is equal to the height of the building. The equation that contains the variables is          = –100 m + (–123 m)      = –223 m
Paraphrase	The height of the building is 223 m.

Read

Read pages 59 to 61 of your textbook to see examples of problem solving with projectiles directed upwards. Pay careful attention to the analysis for part (a) in both “Example 1.18” and “Example 1.19,” and note the equations chosen in both examples in parts (a) and (b).

Self-Check

SC 5. When an object is thrown upwards, what is its final velocity at the maximum height?

SC 6. When an object that is thrown upwards begins to fall back down, what is its initial velocity for the downward path?

Example Problem 8: An object is thrown upward at 49.05 m/s.

• What is the object’s velocity at 3.00 s?
  
  

  Given
  Required	the velocity after 3.00 s ()
  Analysis and Solution	Choose the positive direction to be upward. The equation, which will have to be manipulated to get , is     = (+49.05 m/s) + (–9.81 m/s2) (3.00 s) = +19.6 m/s
  Paraphrase	The velocity at 3.00 s is 19.6 m/s [upward].

  
  

• What is the velocity after 7.00 s?
  
  

  Given
  Required	the velocity after 7.00 s ()
  Analysis and Solution	Choose the positive direction to be upward. The equation, which will have to be manipulated to get , is     = (+49.05 m/s) + (─9.81 m/s2)(7.00 s) = ─19.6 m/s
  Paraphrase	The velocity at 7.00 s is 19.6 m/s [downward].

  
  

  The velocity after 7.00 s is –19.6 m/s.
  
  

• The object momentarily stops moving vertically at the maximum height. How long does it take to reach its maximum height?
  
  

  Given
  Required	the time to reach its maximum height (Δt)
  Analysis and Solution	Choose the positive direction to be upward. The equation, which will have to be manipulated to get Δt, is . Remember, you cannot divide vectors. You can only divide the magnitudes of the vectors.
  Paraphrase	The time to reach its maximum height is 5.00 s.

  
  

• What is the maximum displacement?
  
  

  Given
  Required	the maximum displacement ()
  Analysis and Solution	Choose the positive direction to be upward. The equation to use is   = (+49.05 m/s)(5.00 s) + ½(─9.81 m/s2)(5.00 s)2 = +245.3 m + (─122.6 m) = +123 m
  Paraphrase	The maximum displacement of the object is 123 m [upward].

Self-Check

SC 7. Go to page 63 of your textbook and answer question 10 of “1.6 Check and Reflect.” (Hint: Problems like this are most easily solved by taking the downward half of the projectile’s path.)

Module 1: Lesson 5 Assignment

Remember to submit the answer to TR 4 to your teacher as part of your Module 1: Lesson 5 Assignment.

Try This

TR 4. A water balloon is dropped on some unsuspecting sunbathers from a stationary hot air balloon. The water balloon is 353 m above the beach. If the water balloon accelerates downwards at 9.81 m/s2, how long will it take to hit the beach?

Read

How do these formulas tie into the graphs that you studied in the first four lessons? Read from the middle of page 61 to the bottom of page 62 of your textbook.

Self-Check

SC 8. Go to page 62 of your textbook and answer questions (a) and (b) of “Concept Check.”

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