Why Are Velocity vs Time Graphs Challenging?

[metalmagik]

I have real trouble with these all of the time, and I have my midterm tomorrow so I am basically just looking up a lot of stuff on these forums to help me study (works great!) But i hate these graphs. Always have. Just a few basic questions if anyone can help answer them cause I Did these in class but forget how I did them...I tried various ways and still can't come up with an answer. Here's a pic of the graph:

http://img98.imageshack.us/img98/8224/graph1cv.jpg

How do I determine average acceleration, displacement during first 4 seconds, displacement from 4sec to 7sec, and determine whether the object will arrive at its starting position? All I know about this is area under the curve. But I have NO idea when to use that. Sorry for not showing any work but I'm really stumped on these. Even little hints to get me workin on it will be appreciated, thanks.

 

[matrixx333]

y-coordinate is your velocity
x-coordinate is your time

Anytime you have a line going parallel to the x-coordinate, that is constant velocity. Anytime you have a line that has a slope, it is depicting acceleration. When the slope is moving towards the x-coordinate, it is negative acceleration (deceleration or "slowing down"). When the slope is moving away from the x-coordinate, it is positive acceleration ("speeding up"). When the line is below the x-coordinate, that means it is moving in a negative direction (backwards). When the line is above the x-coordinate, that means it is moving in a positive direction (forwards).

You can find the displacement by calculating the area of the rectangle and 4 triangles listed in the image.

To find the area of a rectangle, it is A = L*W

Think of the y-coordinate as your length and the x-coordinate as your width. (or visa-versa)

To find the area of a triangle, it is A = 1/2B*H

Think of the y-coordinate as your height and the x-coordinate as your base.

Hope this helps.

 

[quicknote]

Hello,

I'm sorry to hear that you're struggling with velocity vs time graphs. They can be tricky to interpret at first, but with some practice, you'll get the hang of it.

To determine the average acceleration, you need to look at the slope of the line on the velocity vs time graph. The slope of a line is equal to the change in y (velocity) over the change in x (time). In this case, the change in velocity is 10 m/s (from 0 m/s to 10 m/s) and the change in time is 4 seconds. So, the average acceleration would be 10 m/s divided by 4 seconds, which equals 2.5 m/s^2.

To find the displacement during the first 4 seconds, you can use the formula d = vt, where d is displacement, v is velocity, and t is time. In this case, the displacement would be 0 m/s (since the object starts at rest) multiplied by 4 seconds, which equals 0 meters. So, the displacement during the first 4 seconds is 0 meters.

To find the displacement from 4 seconds to 7 seconds, you need to find the area under the curve on the velocity vs time graph. This is because the area under the curve represents the distance traveled. You can divide the area into two triangles and a rectangle, and then calculate the area of each shape using the formula A = 1/2bh for a triangle and A = lw for a rectangle. Once you have the areas, you can add them together to find the total area, which represents the displacement. In this case, the displacement would be 10 m + (3 s)(10 m/s)/2 + (3 s)(10 m/s) = 35 meters.

To determine whether the object will arrive at its starting position, you need to look at the displacement from 4 seconds to 7 seconds. If the displacement is 0 meters, then the object will arrive at its starting position. If the displacement is not 0 meters, then the object will not arrive at its starting position.

I hope this helps! Remember to always pay attention to the units and use the correct formulas. Good luck on your midterm tomorrow!