# Homework Help: Acceleration vs Time Graph

1. May 1, 2012

### ev_physics

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

I'm practicing for my exam and so I'm studying off practice tests that my professor gave me.
One of the problems is an acceleration vs time graph. The question is question #2 on the link below.

http://facultyfiles.deanza.edu/gems/lunaeduardo/Physics50Exam1W11.pdf

2. Relevant equations

I'm trying to use
Velocity = acceleration*time
but I'm not given an acceleration

3. The attempt at a solution

I'm assuming because at t = 0s, velocity = 10 m/s.
I understand that Acceleration is velocity (m/s) per 1 second but if velocity at 0s is 10m/s, how do you use that to calculate for a time at maybe, 1s or 2s?
Any help will be great.
Thanks

2. May 1, 2012

### BruceW

hi there, welcome to physicsforums :)

I've looked at the graph, and I think they made a mistake. The y axis of the graph should have values for the acceleration. As it is now, I don't think you can calculate anything from it!

3. May 1, 2012

### ev_physics

Thank you!

Ya, I've been looking at the graph for a long time and trying to figure out why I can't get an answer. At first, I thought it was probably me just not understanding how a A vs. T graph works but after an hour of looking around and watching youtube videos and everything, I concluded that I don't think I can be that clueless...

However, I feel like the "t=0s, then velocity = 10m/s" is where you are supposed to start at maybe?
Its just hard to believe that my professor would give us an incorrect problem to study from for an exam

4. May 1, 2012

### ehild

Maybe your professor meant that one scale on the vertical axis corresponded to 1 m/s2 acceleration. Try to solve it that way. And remember: acceleration is change of velocity over a time interval Δt: a=ΔV/Δt. In case of constant acceleration, V=V(0)+at, but the acceleration varies with time in the figure. In that case, the area under the a(t) diagram gives the change of velocity. For example, the acceleration-time graph encloses a triangle between 0 and 25 s. The height is 7 scales=7 m/s. The "area" of the triangle is [25(s) * 7 (m/s2)]/2= 87.5 m/s. It is the change of velocity between t=0 and t=25 s. To get the actual velocity, add 10 m/s.

ehild

5. May 1, 2012

### BruceW

possibly too much help there. But I do agree, the professor probably just forgot to mention what the scale on the vertical axis was meant to be, so you can still try the problem by taking a guess at what the scale is meant to be. (But if your professor has used a different scale, then your answer may not turn out the same as his, but it is good practice anyway).

6. May 1, 2012

### ehild

Bruce, this is a very unusual problem for students unfamiliar with calculus and just studying motion with uniform acceleration. I do not think I gave too much help by explaining how to relate area under acceleration to change of velocity.

ehild