How Do You Calculate Displacement from a Velocity-Time Graph?

[kmiller]

displacement problems...help!

Homework Statement

A person on horseback moves according to the velocity-versus-time graph shown in Figure 2-29. (The vertical axis is marked in increments of 2 m/s and the horizontal axis is marked in increments of 4 s.) Find the displacement of the person for each of the segments A, B, and C.

Find
m (A)
m (B)
m (C)

 

[LowlyPion]

Homework Statement

A person on horseback moves according to the velocity-versus-time graph shown in Figure 2-29. (The vertical axis is marked in increments of 2 m/s and the horizontal axis is marked in increments of 4 s.) Find the displacement of the person for each of the segments A, B, and C.

Find
m (A)
m (B)
m (C)

Welcome to PF.

What are you stuck on?

 

[kmiller]

I don't know how to get the displacement I tried...doing for A 2m/8s=.25m but it was wrong...what do you add or minus from what?

 

[LowlyPion]

I don't know how to get the displacement I tried...doing for A 2m/8s=.25m but it was wrong...what do you add or minus from what?

Distance is not Velocity/time. What you found was acceleration.

Distance is Velocity times time.

 

[DrDan]

The displacement is velocity * time which is effectively the area under the curve for each segment. For segment A you want to find the area of the triangle under the line marked A. The area of a triangle is ½ the base * height. So you have 0.5 * (8s) * (2m/s) = 8 meters.

Thinking about segment A, the average velocity is 1m/s, the velocity change is linear, and the person on horseback is moving for 8 seconds, so they should move 8 meters.

For segment B you have two pieces to the area under the curve. In the attached picture, The first is the red triangle shown 0.5* (4s)*(6m/s-2m/s)= 8 meters. The second is the blue box shown which is 4s*2m/s = 8m. 8m+8m = 16 m. The answer to part B is 16 meters.

Thinking about segment B, the average velocity is 4m/s, the velocity change is linear and the person on horseback is moving for 4 seconds. So 4m/s*4s is again 16 meters.

The displacement in segment C is twice what it is for segment B. It’s the same average velocity, but for twice the length of time. Segment C displacement is 32m. If you want to do it out you will have to add a triangle plus a rectangle to get the area under the curve.

If the curves are more complex (not just simple lines), you may have to use Integral Calculus to find the area under the curve.

DrDan