I need to find a velocity for a given t value, from a linear acceleration graph

[Andrew22]

Homework Statement

A particle move along the x-axis, each starting with V = 10 m/s at t = 0 seconds.
Find each particle's velocity at = 7.0 . Work with the geometry of the graphs, not with kinematic equations.

Homework Equations

A(t) = {10t +10, 0 ≤ x ≤ 2
-10t+30, 2 < x ≤ 8}
I think I wrote that right, I'm not very good at writing these on the computer yet.
There is a corner at A (t) = 30, which is (30,2) and the lines are connected. (wish I could draw a graph, can't link the picture.)

The Attempt at a Solution

I thought that if I just took the antiderivatives of those equations, evaluated the first equation for t=2 that would give me the velocity for t=2, this gave me 50 (5t^2 +10t +10). I then did the same process using the new C value for the antiderivative of the second equation (50) which ended up (-5t^2 +30t +50), I evaluated this for t=5, as that was the change in T from t=2 to t=7. This all gave me -25, and that was wrong. I have never worked with an acceleration that wasn't a constant. So I am more or less figuring this out on my own. I can't find this in my course text, and google is not coming up with much, probably just entering the wrong things. Any help is appreciated... Am I on the right path?

 

[tiny-tim]

Welcome to PF!

Hi Andrew22! Welcome to PF!

Sorry, but you're completely missing the point of this question …

Work with the geometry of the graphs, not with kinematic equations.
…
I thought that if I just took the antiderivatives of those equations …

No, you're supposed to just read it off the graph …

this is to give you practice at reading graphs.

This graph is an upside-down V, isn't it?

(is A(t) the distance the speed or the acceleration? I'm assuming it's the distance)

ok, how do you tell the speed from just looking at the graph?​

 

[Andrew22]

A(t) is the acceleration, I was unclear about the antiderivative I took working backwards to try and get velocity. But I'm not sure how to see velocity from an acceleration that isn't constant.

And yes it is an upside down V.

I did not use kinematics for this. I was only working with the equations for the slopes.

However, I don't doubt I'm missing the point.

I know how the general shape of the velocity graph will look, but I am unclear as to how to get actual values by just seeing it.

 

[tiny-tim]

Hi Andrew22!

I did not use kinematics for this. I was only working with the equations for the slopes.

Sorry, but the question specifically tells you to use geometry.

Geometry does not include coordinate equations …

Work with the geometry of the graphs, not with kinematic equations.

A(t) = {10t +10, 0 ≤ x ≤ 2
-10t+30, 2 < x ≤ 8}

A(t) is the acceleration …

And yes it is an upside down V.

You have an acceleration vs time graph.

You need the speed, which is the integral, sooo …

how can you read an integral off a graph (at least, an easy straight-line graph like this one)? ​

 

[Andrew22]

Got it thanks, I had not known the integral was the area under the curve. But my proffessor filled me in.