Investigation 1

Units, Linear Motion, Freely Falling Bodies

The following questions (1 through 5) deal with linear motion (motion in a straight line) where the object is traveling at a constant velocity (constant speed). The relationships between distance traveled, speed of travel, and time are:

distance = (constant or average speed) x time

(constant or average speed) =

time =

1.A car is traveling at a constant 30 miles/hr on Pico Boulevard toward the ocean.

a.What is the speed of the car?

30 miles/hr

b.What is the velocity of the car?

30 miles/hr toward the ocean

c. Which quantity is a scalar and which one is a vector. What is the difference between a scalar and a vector?

Speed is a scalar – magnitude only.

Velocity is a vector – both magnitude and direction.

2.A truck is traveling on Ocean Park Boulevard toward the ocean.

a.If the truck is traveling at a constant speed of 20 m/s, then how far will the truck travel in 6 seconds?

Distance = (constant speed) x time = (20 m/s) x (6 sec) = 120 m

  1. If the speed of the truck is changing, but its average speed is 20 m/s, then how far will the truck travel in 6 seconds?

Distance = (average speed) x time = (20 m/s) x (6 sec) 120 m

3.a.A physics student runs a distance of 100 m in 20 seconds at a constant speed. At what speed was the student running?

Speed = (distance)/(time) = (100 m)/(20 sec) = 5 m/s

bA second student starts from rest and then slowly increases her speed until she travels a total distance of 100 m The total time she takes is 20 seconds. Was she traveling at a constant speed? What was her average speed?

Speed was not constant – it changed!

Average speed = (distance)/(time) = (100 m)/(20 sec) = 5 m/s

Note that even though the second student was not traveling at a constant speed, her average speed is the same at the first student’s constant speed, and therefore they both travel the same distance in the 20 seconds.

4.The sketch shows a ball rolling at constant speed along a level floor. The ball rolls from the first position shown to the second position in 1 second. The two positions are 2 feet apart. Carefully sketch the position of the ball at successive 1-second intervals all the way to the wall (neglect any resistance to the motion).

The ball is traveling with a speed of ______2 ft/sec______when it reaches the wall, and takes a

time of approximately ______5 sec______.

5.The table shows data of the average sprinting speeds of some animals. Make whatever computations are necessary to complete the table. Be sure to include the units.

Animal / Distance / Time / Average Speed
Cheetah / 75 m / time = (75 m)/(25 m/s) = 3 sec / 25 m/s
Greyhound / 160 m / 10 s / speed = (160 m)/(10 s) = 16 m/s
Gazelle / 1 km / time = (1 km)/(100 km/hr) = 0.01 hr / 100 km/hr
Turtle / distance = (1 cm/s) x (30 s) = 30 m / 30 s / 1 cm/s

The following questions (6 through 8) deal with accelerated motion. Remember that when the acceleration is constant, the average speed can easily be calculated. That is, when the acceleration is constant:

average speed =

Remember how distance, average speed, and time are related to each other:

distance = average speed x time

6.A student driving a car starts from rest and accelerates at a rate of 2 m/s2 for 10 seconds.

When you answer the following questions, note down (for yourself) how and why you performed the steps that allowed you to arrive at your answer. This will help you remember how to solve the problem.

  1. What is the initialspeed of the car?

Initial speed (starting speed) = 0 (at rest)

  1. What is the speed of the car at the end of the 10 seconds?

The speed of the car changed by 2 m/s for each of the 10 seconds it was accelerating. At the end of the first second: 2 m/s, the second second: 2 + 2 = 4 m/s, the third second: 4 + 2 = 6 m/s,and so on to the tenth second: 18 + 2 = 20 m/s.

c.What is the average speed of the car during the 10 seconds?

Average speed = (initial speed + final speed)/2 = (0 + 20 m/s)2 = 10 m/s

d.How far does the car travel in the 10 seconds?

Distance = (average speed) x (time) = (10 m/s) x (10 sec) = 100 m

7.In the previous problem, the student driving the car now slows down (decelerates) and stops in 20 seconds.

  1. What is the initialspeed of the car?

Initial speed = 10 m/s

  1. What is the speed of the car at the end of the 20 seconds?

At the end of the 20 seconds the car is stopped, so the final speed is 0.

c.What is the average speed of the car during the 20 seconds?

Average speed = (initial speed + final speed)/2 = (20 m/s + 0)/2 = 10 m/s

d.How far does the car travel in the 20 seconds?

Distance = (average speed) x (time) = (10 m/s) x 20 sec) = 200 m

The following question is again linear motion with a constant acceleration, but this time in the downward direction. (Refer to Questions 6 and 7.)

The EXTRA CREDIT question in the last column is difficult only because the ball changes direction. In order to figure out the answers remember first, that the upward and downward motions are symmetric. Second, if the sign for the speed is positive in one direction, then then the sign becomes negative when traveling in the opposite direction.

8.In this question you will examine a freely falling body, i.e., the effect of air resistance on the object is ignored. In each case, the time starts from zero when the ball is dropped or thrown. Take the acceleration due to gravity to be 10 m/s2 (or 32 ft/s2). Make sure you know what you’re doing and why.

EXTRA CREDIT

The ball is dropped / The ball is thrown downward at 100 m/s / The ball is thrown upward at 20 m/s
How fast (initial velocity) and in what direction is the ball traveling at 0 sec? / 0 m/s / 100 m/s, downward / 20 m/s, upward
How fast and in what direction is the ball traveling at 1 sec? / = 0 + 10 m/s =
10 m/s, downward / = 100 m/s + 10 m/s = 110 m/s, downward / 10 m/s, upward
What is the average speed from 0 to 1 second? / 5 m/s / = (100 m/s + 110 m/s)/2 = 105 m/s / = (20 m/s + 10 m/s)/2 = 15 m/s
How far (distance) is the ball from its starting position at 1 second? / = (5 m/s) x (1 sec) =
5 m / 105 m / = (15 m/s) x (1 sec) = 15 m
How fast and in what direction is the ball traveling at 2 sec? / 20 m/s, downward / = (110 m/s + 10 m/s) = 120 m/s / = +10 m/s – 10 m/s = 0
What is the average speed from 0 to 2 seconds? / = (0 + 20 m/s)/2 =
10 m/s / 110 m/s / = (20 m/s + 0)/2 = 10 m/s
How far (distance) is the ball from its starting position at 2 sec? / = (10 m/s) x (2 sec) = 20 m / = (110 m/s) x (2 sec) = 220 m / 20 m above
How fast and in what direction is the ball traveling at 4 sec? / = 40 m/s, downward / 140 m/s, downward / = +20 – 10 – 10 – 10 – 10 = - 20 m/s
What is the average speed from 0 to 4 seconds? / = (0 + 40 m/s)/2 =
20 m/s / = (100 m/s + 140 m/s)/2 = 220 m/s /
How far distance) is the ball from its starting position at 4 sec? / 80 m below / = (220) x (4 sec) =
880 m / = (0) x (4 sec) = 0 m, same height as thrown

*This downward speed is written as a negative value.