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Uniform Motion Model

Page history last edited by 25oliviag@students.harker.org 2 years, 2 months ago Saved with comment

Uniform Motion Practice Problems

 

In both physics and everyday life, "uniform" means "the same." For uniform motion, the velocity is uniform or constant. This means the object must travel in a straight line at a constant speed to be traveling uniformly. An animation for a box moving uniformly can be found here. The purpose of our Uniform Motion model is describe what happens when an object moves uniformly. We use graphs, diagrams (i.e. motion maps), words, and algebra (i.e. uniform motion math model) to describe Uniform Motion. 

 

Example of Uniform Motion

 

Scalars vs. Vectors

Knowing when direction is included is one of the most important components of physics. If you don't include direction in values that are vectors you will get points off!

  • Scalars are quantities that only have magnitude (or size). 
  • Vectors are quantities that have magnitude and direction. A direction should always be written in words in a final answer, but it's ok to use "+" and "-" in an equation. 

 

Example:

 

Write five variables which are scalars:

Answer: time, mass, distance, volume, & temperature

 

Write five variables which are vectors:

Answer: force, velocity, acceleration, position, & displacement. 

 

 

Position & Displacement

 

Terms to know:

  • d = position (NOT distance!!!!!)
  • ∆ (delta) = change in __ (for example change in time = Δt = tf - ti
  • ∆d = displacement (the change in position: final position - initial position)
  • t = time
  • Speed = total distance/total time (does not have direction) 
  • v = velocity [NOTE: not speed, for it has direction (Speed vs. Velocity Song - YouTube)]

 

 

Position vs. Time Graphs

 

 

Position vs. time graphs tell us the position of the object at each location. Based on the positions, we can infer the motion of the object. For example, the position vs. time graph to the left shows an object which starts to the left of the origin and moves to the right at a constant velocity for some time. It then stops for a short amount of time and finally continues to move to the right at the same constant velocity for a longer amount of time. 

 

 

Velocity vs. Time Graphs

 

 

A velocity vs. time graph tells us the velocity of the object at each instant. Since velocity is a vector we can tell which way the object is moving from the velocity vs. time graph. Additionally the graph shows how fast the object is moving at each instant in time. You can use the slope of a position vs. time graph to create a corresponding velocity vs. time graph. 

 

For the velocity vs. time graph to the left, the object starts at an unknown position. Then the object moves to the right at a constant speed for a period of time, stops for a period of time, and continues to move to the right at the same constant speed. 

 

In velocity vs. time graphs, where the object starts does not matter (keep in mind there is no way to determine the position of an object by looking at a velocity vs. time graph for it only shows the direction it is going along with the acceleration). When drawing a velocity vs. time graph, it is not necessary to connect separate horizontal lines. If the velocity was negative, then you would draw the velocity on the bottom half of the y-axis and underneath the x-axis. If the velocity is 0 m/s the entire time, make sure to draw it on the t-axis, so it shows that there is no velocity. If you are not given exact numbers for velocity, use common sense and your intellect to the base where the lines should be (Basically it is ok to approximate as long as comparisons are correct). If you are given numerical values, be sure to label both the axes, and include units. This computer model shows an animation of a moving object and then creates the position vs. time and velocity vs. time graphs for the motion. 

 

 

Written Descriptions of Uniform Motion

 

Written descriptions describe the motion of the object in words. When writing this, make sure to always include what side of the origin the object starts at and the direction it travels. If the object has a negative velocity, it is moving to the left (or down), and if it has a positive velocity, it is going to the right (or up). Also, always specify if the object is moving at a constant speed or speeding up. Finally if there are two or more motions you must compare them (i.e. which one is faster?). 

 

 

Motion Maps

Motion maps are diagrams which represent the motion of the object using dots and arrows.

 

Example

 

 

When drawing a motion map, the dots represent the position of the object at each instant in time. Dots always represent a constant increase in time, so the second dot might be t=1 s.  Remember to label the first dot as t=0 s. Additionally remember that if the object is moving at a constant velocity you need to has the velocity vectors. Finally, remember to label at least one velocity vector with a "v." 

 

Verbal description of the motion map:

The object starts to the right of the origin and moves right at a constant speed. It suddenly starts moving at a faster constant speed, while continuing to the right. 

 

This website will create motion maps from position and time data! 

 

Algebraic Descriptions of Uniform Motion

 

Algebra is useful to describe uniform motion when there are messy numbers or multiple objects. 

 

General Math Model for Uniform Motion:

 

Δd=v*Δt      or           df = vΔt + di

 

Example 1

Two bikes start 12 meters apart and move towards each other at the speeds shown below. First, write algebraic equations to describe the motion of each bike. Then use your equations to determine where the bikes collide. 

 

Equation for bike A:

dfA = vAΔtA + diA

dfA = 1.5m/s*ΔtA + 0m

 

Equation for bike B:

dfB = vBΔtB + diB

dfB = -3.5m/s*ΔtB + 12m

 

To meet the two bikes must be at the same position (dfA=dfB) at the same time (ΔtA=ΔtB). So:

dfA dfB   and ΔtAtB

 

dfA = 1.5m/s*ΔtA + 0m = -3.5m/s*ΔtA + 12m

 

ΔtA 

 

dfA = 3.6 m to the right of the origin (they meet) 

 

 

 

Example 2

 

Problem Solving Approach in Physics

 

Multiple Representations

In order to be proficient with Uniform Motion, you should be able to seamlessly describe objects which move at constant speeds using graphs, equations, words, and diagrams. 

 

 

 

Practice Problems:

1. Does measurement of position depend on the choice of coordinate system? Why/Why not?

 

2. Does the measurement of displacement depend on the choice of coordinate system? Why/Why not?

 

3. A race car travels at the speed of 100 m/s. How far does it travel in 12.5 seconds?

 

4. Which of the following are impossible on a velocity vs time graph?

     a. horizontal line

     b. straight line with negative slope

     c. straight line with positive slope

     d. vertical line

 

5. Is acceleration a vector or a scalar? Why or why not?

 

6. If the line of a D vs T graph is straddling the x axis, this means the position is...

 

     a. 0

     b. 1

     c. 2

     d. 0.01

 

 

 

7. Answer the questions for the above graph. 

    1. What is the physical interpretation of the vertical intercept?

Advice: so, physical interpretation means the starting point. At first, personally, I didn’t understand what it was meant by physical interpretation. This basically means the understanding of the physical impact. Based on where it starts of is a huge impact on its position obviously, and vertical intercept is pretty self-explanatory. So overall, just look for the starting position (what side of the T axis it is on, what number (if any) is it on, and how many seconds have passed)

Solution: the physical interpretation of the vertical intercept of the car is 10m to the right of the origin. 

  1. What is the average velocity?

Advice: to find the average velocity, first you should look for what goes where in the formula. The formula for average velocity is V=∆d/∆t. The next step you should take is translating the formula. Well, velocity equals the change in position (not distance!) over the change of time, so the final position - the initial position, and the final time - the initial time. In this case, the final position is 60m at a final time of 6 seconds. The initial position was 10 m at a time of 0 seconds. This would mean ∆d = 60m-10m, and ∆t=6s-0s. When you divide the two, you get 58.3 m/s, and that is how you find average velocity

Solution: 58.3m/s

  1. What is the mathematical equation to describe the motion of the car?

Advice: when you make a mathematical equation with position vs time graphs, you need to base it off y=mx+b. In this case, you start with the final position of the car, because that is what you are trying to find. To find the final position, you would have to multiply the velocity by the change of time and add the initial position to it. So, it would look like df=v∆t+di.

Solution: df=58.3m/s(time)+10m

  1. How fast until the car runs out of gas?

Advice: In order to solve this problem, you just need to look at the graph! If the line in your graph starts going to the left, then the car would most likely be running out of gas. But in this graph, it is at a constant velocity going to the right, so we know that the car will not run out of gas under 6 seconds.

Solution: not under 6 seconds.

  1. Assuming the car has a full tank of gas, where will the car be after 10 seconds?

Advice: To solve this problem, you just need to use the mathematical equation that you came up with in one of the previous problems, but plug in the final time (in this case its 10 seconds)

Solution: df=58.3m/s(10s)+10m=593 meters to the right of the origin. 

 

8. If a student throws a rock off a 10 meter cliff at 2 m/s, how far would the rock travel?

 

SOLUTION:

 

We need to first figure out the time it takes for the rock to hit the ground (which is the vertical velocity). The initial velocity would be 0, because there is no downwards force on the object yet (besides the gravity which is 10 m/s^2). We will then need to use the equation d = ½ a * t^2 + vi * t to find the time (since we already know all the other information).

 

Then you plug in the numbers the question gave, and then the equation would be: 

 

10 m = ½ 10 m/s^2 * t^2 + 0 m/s * t

t^2 = 1s^2

t = 1s

 

Now that we know that it would take the rock 1 second to hit the ground, we can now calculate the horizontal distance that the rock traveled. We can use the equation d = v * t.

 

Then you plug in the numbers, and the equation would be: 

 

d = 2 m/s * 1s

d = 2 m

 

Answer: The rock traveled 2 meters

 

 

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