# Homework Help: Given acceleration graph, determine v vs. t and x vs. t graph?

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1. Oct 12, 2014

### timnswede

1. The problem statement, all variables and given/known data
The title, and assume that t=0, x=0 and v=0.

2. Relevant equations
Kinematic equations, and area of triangle and rectangle.

3. The attempt at a solution

I see what I did wrong with the part that is circled "parabola", as it should be just a straight line, but I am lost as to what is wrong at the part that is circled "curved up". It's speeding up during that part so wouldn't it make sense that the slope is steepest at the top of that section?

2. Oct 12, 2014

### Staff: Mentor

Instead of guessing, why don't you divide up the time between 10 and 25 secs and on your velocity graph "count squares" for each increment in area as you step through time?

And where you have the velocity steady at 0 m/s for a number of seconds, what will be happening to your displacement over that interval?

There's no excuse for guessing when you can measure directly off the graphs!

3. Oct 12, 2014

### timnswede

I think I see what I did wrong, it should go all the way down -375, then a straight line, and then curved down again another 84.375 m right? This was using kinematic equations. So from 10 to 25 seconds it would be curved up, but instead of going towards the positive x it would keep going negative? And when velocity is 0 displacement doesn't change. This is a rough sketch of what it would look like.

4. Oct 12, 2014

### ehild

You are right that the displacement is negative all the time, as the velocity is negative.
But think of the equation for displacement, when the initial velocity is zero: x=a/2 t2. It is a parabola, upward open when a >0 and upside down when a<0. Check the curvatures.

5. Oct 12, 2014

### timnswede

Ah OK, that makes sense. I think I was too caught up with velocity being negative. The way you said I should do it makes it much easier though, thank you.

6. Oct 12, 2014

### Staff: Mentor

Are you going to post another sketch?