# Velocity Vs. Time Graph?

1. Feb 17, 2010

### huybinhs

1. The problem statement, all variables and given/known data

The graph is:

http://i995.photobucket.com/albums/af79/huybinhs/Lastproblem.gif

A particle moves along the x-axis. The velocity of this particle as a function of time is shown in the figure. Assume the particle is located at x = 0 m at time t = 0 s.

1. What is the acceleration of the particle at time t = 26.0 s?

2. What is the position of the particle along the x-axis at time t = 26.0 s?

3. What is the net displacement of the particle between time t = 4.0 s and t = 38.0 s?

2. Relevant equations

x = x0 + v0t + 1/2 at2

v = v0 + at

2(x-x0) = (v+v0)t

v^2 = v^02 + 2a(x − x0)

g = 9.8 m/s^2

3. The attempt at a solution

1. - 0.3 m/s² ANS {by slope of graph at t = 26.0} => incorrect answer.

2. x = (-2)(26) + (0.5)(-0.3)(26)² = -52 - 101.4 = -153.4 ANS => incorrect answer.
{use eq for x at t = 26}

3. - 7.5 m ANS {by area under graph from t = 4 to t = 38} => incorrect answer.

2. Feb 17, 2010

### CompuChip

a) How did you calculate the slope? I got a different answer.

b) The formula x0 + v0 t + 1/2 a t^2 is only valid when a is constant. That is not true between t = 0 and t = 26. In this case, you need to determine the displacement from the area under the graph.

3. Feb 17, 2010

### huybinhs

Now I got these:

a) a = (v-v0)/t (t0 = 0s, v0 = 3m/s; t= 26s, v =-1.75m/s)

=> a = (-1.75 - 3)/ 26 = -0.183 m/s^2 => incorrect.

b) t0 = 0s, v0 = 3 m/s; t = 26s, v = -1.75 m/s

=> (-1.75 - 3) * (26 - 0) = -123.5 m => incorrect.

c) t0 = 4s, v0 = 4.25 m/s; t = 38s, v= 1 m/s

=> (4.25 * 4) + (38 * 1) = 55 m => incorrect.

4. Feb 17, 2010

### huybinhs

Anyone?

5. Feb 17, 2010

### huybinhs

I still couldn't get the right answer!!! Help please!!!

6. Feb 18, 2010

### huybinhs

?????????????

7. Feb 18, 2010

### CompuChip

Please stop making wild guesses and take a deep breath.

You are treating the graph like it's a straight line, which it isn't.

For a, you need the slope of the graph at t = 26, you calculated the slope of the line going straight from (0, 3) to (26, -1.75). You can only accurately calculate the slope for a (piece of) a straight line. Between t = 20 and t = 30 you have such a piece, not between t = 0 and t = 26.

Similarly for the area. You are simply calculating the area as a rectangle, which it is not. Also here, you should divide the time in smaller parts, on each of which the graph is a straight line segment. Then the area you are looking for is a square + a triangle. For example, what is the displacement from t = 0 to t = 10? Well, if you shade the area between the x-as and the graph, you see a rectangle with corners (0, 0), (0, 3), (10, 3), (10, 0) with area 3 x 10 = 30. On top of that you see a triangle with corners (0, 3), (10, 7), (10, 3) which has area 1/2 b h = 1/2 x 10 x 4 = 20. So the area you are looking for is 50.