Converting Velocity-Time Graphs Into Acceleration Graphs

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j1249
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I am completely lost on how to convert velocity time graphs into acceleration time graphs.
so far all i no is to break the velocity graph into sections/intervals and find the area. the answer i get for the area is the scale on my y-axis of my acceleration graph. but what i do not no is what kind of line would it be on the acceleration graph (like would my line be straight and positive or straight and negative or would it be just a horizontal line) and how to determine that? Also how do i convert graphs that go from a positive scale to a negative scale in one graph? any help would greatly be appreciated! Thank you!
 
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j1249 said:
I am completely lost on how to convert velocity time graphs into acceleration time graphs.
so far all i no is to break the velocity graph into sections/intervals and find the area. the answer i get for the area is the scale on my y-axis of my acceleration graph. but what i do not no is what kind of line would it be on the acceleration graph (like would my line be straight and positive or straight and negative or would it be just a horizontal line) and how to determine that? Also how do i convert graphs that go from a positive scale to a negative scale in one graph? any help would greatly be appreciated! Thank you!

This may be your problem.

The areas under your Velocity-time graph are what? Like what are the units? i.e. what are the units of V*t?

What is the definition of acceleration?
 
well the units of my v-t graph is m/s for velocity and s for time. as for the definition i do not no what you are asking. in my question it only asks me to determine the acceleration for each interval. then draw an a-t graph. so for my first interval i found the area by doing bh/2 so (2)(4)/2 and got 4m [forward]
 
j1249 said:
well the units of my v-t graph is m/s for velocity and s for time. as for the definition i do not no what you are asking. in my question it only asks me to determine the acceleration for each interval. then draw an a-t graph. so for my first interval i found the area by doing bh/2 so (2)(4)/2 and got 4m [forward]

The area under a v-t graph is unfortunately distance.

Velocity times time is just that - distance, expressed graphically as an area. Because that is the dimension (m/s)*s = just m.

So if velocity is change in position divided by the change in time, what is acceleration?
 
change in velocity divided by the change in time??

right...? lol