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Lecture Comments (9)

3 answers

Last reply by: Jason Wilson
Tue Dec 23, 2014 11:45 AM

Post by Jason Wilson on December 23, 2014

I need more clarification on this. Isn't the speed of the ball at the "ground" level, in the students hand, zero? Not going up or going down? haha. At the point of release of the ball, is when the energy is transferred?  That is how I am seeing this. As I am typing I can "picture" the ball being fastest at the split second it is released out of the hand, but its initial speed before that happens is zero, at ground level in the kids hand. This is pretty funny to be analyzing this at 4:50 AM. Let me tell you how much I enjoy your lectures, you are a very good teacher. I love this stuff. I really appreciate help on this.

1 answer

Last reply by: Professor Dan Fullerton
Sun Dec 21, 2014 9:17 AM

Post by Jason Wilson on December 21, 2014

Its initial velocity is Zero at the ground? right?

2 answers

Last reply by: Jason Wilson
Tue Dec 23, 2014 3:53 AM

Post by Jason Wilson on December 21, 2014

Number 5 It appears that the graph does not show it starting at zero. If the ball is thrown up from the ground. Isn't the ground. . . ground zero (haha)? The acceleration slows as time increases. The acceleration reaches a point of no acceleration, Then reverses, coming back to its origin, which was the ground,(zero... why is it showing going below its origin?

Motion Graphs

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Question 1 0:13
  • Question 2 2:01
  • Question 3 3:06
  • Question 4 3:41
  • Question 5 4:30
  • Question 6 5:52

Transcription: Motion Graphs

Hi everyone and welcome back to 0000

In this mini-lesson, we are going to go over the APlusPhysics worksheet on kinematics graphing motion and you can download that worksheet at the link below.0002

Let us dive right in.0011

Number 1 -- A cart travels with a constant non-zero acceleration, so we have some amount of acceleration or a change in speed, a change in velocity along a straight line. 0013

Which graph best represents the relationship between the distance the cart travels and the time of travel? 0024

We are being given here distance time graphs and we want the one that shows a constant non-zero acceleration. 0030

Well, if you recall, the way you get velocity from a distance time graph is to take the slope. 0036

If I start over here, I start with a very low slope and I go to a high slope. 0042

That means we must have a change in velocity. 0048

A velocity graph would probably look something like that if it were a velocity time graph. 0050

Here we have a negative constant slope, so we would have a negative velocity. 0056

Here we start out with a positive slope, so we would start up here for velocity, where we have a 0 slope and over here it is negative, so we would probably have a velocity time graph that looked like that. 0061

Over on the right, we have a constant positive slope, so we would have a constant positive velocity, but we want the one with a non-zero acceleration and it has to be constant. 0072

Up here we have a change in velocity, but at a constant rate and the slope of this would give us the acceleration, which would be positive in constant, so purple, if that is our acceleration time graph, is a constant value and it is non-zero, so our correct answer must be 1. 0084

Or as you look at this, you can think about it from the graph that you have a change in distance over time and the rate at which your distance is changing is increasing and it is increasing at a constant rate. 0103

You can use that to solve it too. Regardless, the best answer there is Number 1. 0115

Number 2 -- We have a car on a straight road that starts from rest and accelerates at 1 m/s2 for 10 s, then the car continues to travel at constant speed for an additional 20 s. 0122

Find the speed of the car at the end of the first 10 s. 0132

Well, what we know when it starts out is it starts from rest, so our initial velocity is 0 and it accelerates at 1 m/s2 for a time of 10 s. 0135

For the first 10 s, we want to know the speed at the end of that first 10 s. 0150

Well, final velocity equals initial velocity plus acceleration times time, so that is going to be 0 + 1 m/s2 (acceleration) × 10 s (time)...0155

...which implies then that our velocity is going to be 1 m/s2 × 10 s is just going to be 10 m/s. 0169

The speed of the car at the end of the first 10 s is 10 m/s. 0178

Continuing on with the same story, a car on a straight road has the same story, but on the grid below, we are going to use a ruler to construct a graph of the cars speed as a function of time for the entire 30 s interval. 0185

At the end of 10 s, we know that our car was going 10 m/s, so I can draw that on my graph and then it continues to travel at constant speed for an additional 20 s, so that would be our graph of speed versus time. 0197

Number 4 says calculate the distance the car travels in that first 10 s and there are a couple of ways we could do that. 0222

You could take the graph and you could take the area under the graph or you could use your kinematics. 0229

Since we are talking about motion graphs, let us take the area under the graph in that first 10 s of the graph. 0233

And the shape there is that of a triangle, so we had 10 s here, we had a maximum speed of 10 m/s, so the area then of our triangle, which will give us the distance traveled... 0238

...The area of a triangle is 1/2base × height, so that will be 1/2 × 10 s × 10 m/s or 50 m and you could have also have done that with kinematic equations as well. 0252

Number 5 -- A student throws a baseball vertically upward and then catches it when it comes back down. 0269

If we consider vertically upward the positive direction -- so up is positive (y) -- which graph represents the relationship between velocity and time for the baseball, neglecting friction? 0277

When I throw it up, the moment it leaves my hand, it has a large positive velocity and as it goes further and further up, it slows down, slows down, slows down and at its highest point it stops, so part way through its path it is going to have 0 velocity and then it is going to come back down. 0288

It has a greater and greater magnitude, greater and greater speed, but in the opposite direction, so it is going to have a larger and larger negative velocity. 0305

The graph that shows this best is Number 4. 0311

We could also take a look at this graph and say if we wanted to know the acceleration, what would the acceleration look like? 0316

The acceleration is going to be the slope of that line, which the slope of that line is negative and constant, so our acceleration time graph would be down here at negative. 0322

That is actually going to be -9.8 m/s2 as the acceleration due to gravity. 0334

What would that negative mean? Well, if we called up the positive direction, the acceleration due to gravity is in the opposite direction down, so that negative is indicating the direction there. 0341

Let us take a look at one more. The graph below represents the displacement of an object moving in a straight line as a function of time. 0351

What was the total distance traveled by the object during this 10 s interval?0359

Let us take a look and see what we can figure out. 0365

As we start here at (0,0) and we go up through that distance, we have traveled 8 m .0367

Then we have no change in displacement there and the next 2 s, we travel from 8 to 16 m, so we travel another 8 m. 0375

Then, finally, we come back another 8 m, but because it is distance traveled, that is the total amount of distance you are covering, so we have to add that in. 0388

Our total would be 24 m and our best answer there is Number 4. 0398

That finishes page 1 of the worksheet. 0404

Thanks so much and make it a great day everyone!0406