Finding Displacement on a Velocity Time Graph

1. Oct 31, 2003

JDK

Hello all,

I'm pretty well a newbie to physics and need a little help. It has to do with velocity time graphs. Here is the question I am having trouble with...

Time (s) Velocity (m/s)
0.0 0.0
1.0 5.0
2.0 4.0
3.0 10.0
4.0 15.0
5.0 18.0
6.0 20.0
7.0 22.0
8.0 25.0

a) Plot a velocity time graph

<-- this was simple. I had no problems with this component. Time on the x-axis and velocity on the y-axis. Plot the ordered pairs. Draw straight lines connecting the points.

b) Determine the distance the car travels during the first 2.0 s

<-- Here's where I had problems. In the book I am using ... (Merril Physics: Principles and Problems)... its examples only include graphs which are linear and none which produce scattered plots. (or at least from what reading was assigned for this worksheet)If anyone would be so kind to explain to me how to do this type of question with non-linear graphs I'd be grateful. I've been somewhat frustrated with it of late. I know I have to find the area of the space under the curve within the specific time interval but from this graph, 0-2s creates a shape which is not a triangle or rectangle... or any shape which i know has a formula for area.

- The Guy Who Needs Some Help In Physics

2. Oct 31, 2003

jcsd

You'd normally inetgrate the function within the limits to find the area under a graph. The problem here though is that you don't have the equation of the line and the accelration isn't even constant.

The best approacgh to this problem would probably be to divide the area under the line into strips and treat each strip as a rectangle and work it out by adding htere areas.

3. Oct 31, 2003

jcsd

Actually, perhaps the area under the limits you want is to small to divide up, if that's the case you could use a line of best fit.

4. Oct 31, 2003

HallsofIvy

What the values you give:
Time (s) Velocity (m/s)
0.0 0.0
1.0 5.0
2.0 4.0

do make is a triangle and a trapezoid. The line from (0,0) to (1, 5) forms a triangle with the lines y=0 and x= 1. You can easily find its area. The line from (1,5) to (2, 4), together with the lines y= 0, x= 1 and x= 2, forms a trapezoid. The two lines x= 1 and x= 2 are parallel and, so, are the bases- they have lengths 5 and 4 so the "average base" is 9/2. The height is the distance from x= 1 to x= 2: that's 1. The area of the trapezoid is (9/2)(1)= 9/2. Now add that to the area of the triangle to find the total area- the distance traveled.

5. Oct 31, 2003

JDK

Thanks so much for your help jcsd and HallsofIvy. I was considering doing it the way you described but thought I was wrong. Guess sometimes I should at least try my intial thoughts out. Thanks again.