Change in velocity when given a graph that shows time and acceleration

[flynnk567]

Homework Statement

The graph in the figure describes the acceleration as a function of time for a stone rolling down a hill starting from rest.
Find the change in the stone's velocity between 2.5s and 7.5s.

Homework Equations

I think v = v0t +1/2at^2

The Attempt at a Solution

I honestly have no idea about how to solve this problem. My professor gives us mastering physics homework problems before he teaches us how to do the material and I'm completely lost right now :/ Any help would be appreciated.

 

[PeterO]

Homework Statement

The graph in the figure describes the acceleration as a function of time for a stone rolling down a hill starting from rest.
Find the change in the stone's velocity between 2.5s and 7.5s.

Homework Equations

I think v = v0t +1/2at^2

The Attempt at a Solution

I honestly have no idea about how to solve this problem. My professor gives us mastering physics homework problems before he teaches us how to do the material and I'm completely lost right now :/ Any help would be appreciated.

The change in velocity is the area under the a-t graph. You need to draw the two vertical ordinates in - at 2.5 and 7.5 - to enclose a trapezium; and calculate that.

 

[Wolvenmoon]

I come from Google!

If you draw out the graph you might notice that your equation a_x = 1*t+2. Drawing lines at x=4.5 and x=7.5 gives you a_x values of 6.5 cm/s^2 and 9.5 cm/s^2. The area of a trapezium can then be calculated by the integral of (1t+2)dt from the lower bound 6.5 to the upper bound 9.5.