Look at the spacing between these cars on this motion diagram.
The spacing between the cars is the same, so that tells us that the motion is the same. This car is moving with a constant, rightward (+) velocity of 10 m/s.
Here is the velocity time graph that represents the motion diagram. Notice that the velocity is 10 m/s for the entire graph. When a velocity time graph is a horizontal line, the velocity is constant. There is no acceleration.
Looking at the spacing between the points on a position or distance vs time graph can help us describe the motion.
When we studied distance time graphs, we looked at the spacing between the points and that helped us describe the motion. On a velocity time graph, we need to think about our speedometer in our car. Look at the velocity points (the y-axis) and think if those were the numbers on your speedometer. How would you be driving? If you were going 10 mph, 10 mph, 10 mph, 10 mph, then you would be moving, but not speeding up or slowing down. You would be moving with a constant velocity and therefore would have no acceleration.
Look at the spacing between these cars on this motion diagram.
The spacing between the cars is getting larger with each second that passes. This tells us that the car is moving with a CHANGING, rightward (+) velocity. The car is moving to the right and speeding up with a constant acceleration.
Look at this velocity time graph. If you think of the velocity points as numbers you see on your speedometer, then your speedometer is readings are 10 m/s, 20 m/s, 30 m/s, 40 m/s, and then 50 m/s. The car is moving with a increasing, positive velocity. You have a constant acceleration.
Positive velocity with zero acceleration.
Positive velocity with a positive constant acceleration.
Look at the pictures above. The two graphs on the left are in the positive velocity region. The area above the x-axis. It doesn't matter if the slope is positive or negative, if the line is in the positive velocity part, then the object has a positive velocity. Remember, having a positive or negative velocity tells us what direction the object is moving. If the object is has a positive velocity, then it is moving to the right or down.
The two graphs on the right are in the negative velocity region. The area below the x-axis. It doesn't matter if the slope is positive or negative, if the line is in the negative velocity part, then the object has a negative velocity. It is moving to the left or up.
Look at the velocity time graph on the the left. Both of those objects are speeding up. If you think about your speedometer again, both of those lines are moving away from zero. In your car, if your speedometer is moving away from zero, you are speeding up. The line above the x-axis is speeding up in a positive direction and the line below the x-axis is speeding up in a negative direction.
The graph on the right shows the object slowing down. Both of the lines are moving towards the x-axis, towards a velocity of 0 m/s. If your speedometer was moving toward zero, you would be slowing down. The line above the x-axis is sslowing down in a positive direction and the line below the x-axis is slowing down in a negative direction.
If an object changes direction, the velocity time graph will show the line crossing the x-axis. In order to change directions, an object has to slow down, stop, then change directions.
In this velocity time graph, you see that the object started with a certain velocity walking to the right. This object was slowing down because the line is moving towards the x-axis. When the line is on the x-axis, the object has stopped. It then changed directions when it moved below the x-axis, into the negative velocity part. The object is now walking to the left and speeding up.
For the four questions below, think about the question, then click on the down arrow when you have your answer to check to see if you are correct.
What is the equation to find slope?
m=∆y/∆x
2. What is the equation to find acceleration?
a=v/t
3. What is the variable on the y-axis on a velocity-time graph?
Velocity
4. What is the variable on the x-axis on a velocity-time graph?
Time
In the questions above, you answered some important topics. You said that slope is ∆y/∆x. You said that a=v/t and that velocity is on the y-axis and time is on the x-axis.
If you look at the equation, you can see that m=∆y/∆x. ∆y is in the numerator, and since velocity is on the y-axis, we will place v in the numerator as well. ∆x is in the denominator and time is on the x-axis, we will place t in the denominator as well.
You can now see that the slope of a velocity time graph is v/t, which is acceleration.
For the nine questions below, think about the question, then click on the down arrow when you have your answer to check to see if you are correct. For questions 1-3 use the animation below to answer the questions.
Which car (red or blue) has a constant velocity the entire time? What is the car's velocity?
The blue car has a constant velocity of 10 m/s.
2. What is happening to the red car from 0-3 seconds?
The red car is speeding up in a positive direction from 0-3 seconds.
3. What is happening to the red car from 3-12 seconds?
The red car is traveling at a constant velocity of 12 m/s from 3-12 seconds.
For questions 4-9, consider the graph below. True or false, is the object...
4. True or false, is the object moving in a positive direction?
True. The line is in the positive region.
5. True or false, is the object moving with a constant velocity?
False. The line is not horizontal.
6. True or false, is the object moving with a negative velocity?
False. The line is in the positive velocity region. It is above the x-axis.
7. True or false, is the object slowing down?
True. The line is approaching the x-axis.
8. True or false, is the object changing directions?
False. It never crossed the x-axis.
9. True or false, is the object moving with a constant acceleration?
True. The line is diagonal.
Displacement from a velocity-time graph can be found by finding the area under the curve.
This velocity time graph shows a constant velocity of 8 m/s. If I wanted to see how far the object traveled from 1-3 seconds (which is the part shaded in green,) all I have to do is find the area of the green shaded part.
Area of a rectangle is length times width, so the displacement is 2 seconds times 8 m/s, which would be 16 m. This makes sense. If I traveled 8 m/s for 2 seconds, I would have moved 16 m.
This works because of math. v=d/t and if I manipulate the equation and solve for distance, I get d=vt. When I found the area of the rectangle above, I multiplied the velocity by the time. Same as d=vt.
You can find the area of an irregular shape (the purple one) as well. You would just have to break up the purple shape, into two shapes that you can find the area of - the red rectangle (l x w) and the blue triangle (Area = (1/2) l x w.) Once you knew the area of the red rechange and the blue triangle, you would just have to add them together to find the displacement that an object moved from 1-3 seconds on the purple shaded area.
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