What does the area under acceleration-time graph indicate?

Question

Some people say it's average velocity while others say it's change in velocity, so what's the right answer?

According to this site it's the change in velocity...
http://id.mind.net/~zona/mstm/physics/mechanics/kinematics/slopesAndAreas/areaOfavst/areaOfavst.html

Answer 1

To fully understand what the website you posted is trying to say, you'd have to know a little bit about Calculus. An area under a curve is an integral. You have to first understand derivatives and integration before this can make sense. But if you already know all this, then all I'm doing is refreshing your memory, otherwise, undestanding area under a curve will take a few lessons to fully understand. But in a geometrcial explanation, you take the first derivative of the position function and multiply it times the size of each sub-interval within your main interval, then you sum all the products of these sub-intervals within your main interval. (Riemann Sums). this will give you the area under the curve.
when given a position function, you can take the first derivative a.k.a velocity, the second derivative is acceleration.
The website you posted is just trying to demonstrate a proof that explains how that the area under a position function within a certain interval is the equivalence to velocity.

Answer 2

It's velocity. Acceleration is change in velocity by definition. I don't know if you've taken calculus, but the integral is the area beneath a curve, and the derivative is the slope of the line at any given point. Integration and differentiation are opposites. If you integrate velocity (m/s), you get distance traveled (m), and if you differentiate velocity, you get acceleration (m/(s^2)). If you integrate the acceleration, you get the velocity, as this is the opposite of differentiating the velocity to get acceleration. It's not the average velocity, it's the cumulative velocity. The idea is that as something accelerates, it constantly changes speed, and if you sum up all of those changes, you get the net result. Mathematically, that's the same as the area under the curve.

Answer 3

It is the change in velocity!
1. Proof by definition
Definition of acceleration is given by change in velocity per time:
(v-u)/t
For every second, the velocity increases by delta v(change in velocity)
For a constant acceleration a, the change in velocity is same for every second or any time unit!
Therefore the total change of velocity is given by the area of an a-t graph
***Note that a constant velocity has 0 acceleration AND we CANNOT know the initial velocity from an a-t graph!

2. proof by v=u+at
The velocity after a period of time is given by the formula above,
v=u+(v-u) and v=u+at
Therefore,
at=v-u
at=CHANGE in velocity

Hope this helps

Answer 4

If the acceleration is equal a constant (horizontal line at h=a) then the area between t1 and t2 is
Area=(t2-t1)h but
v1=v0 +at1 and v2=v0+at2 so
v2-v1=a(t2-t1) where a=h.
which is the change in velocity
BUT
if you calculate the area from 0 to time t then the area will be
Area =ht or setting h=a
Area =at
which is the velocity at time t if the initial velocity was zero.
This is true even if acceleration is not constant but you need calculus to prove it. IT IS NOT AVERAGE VELOCITY IN EITHER CASE.

Answer 5

“Acceleration x time” is “replace in velocity x time / time”. for this reason the for any acceleration time graph the realm provides the classic replace in velocity for the period of that element.

Answer 6

Area Under Acceleration Time Graph

Answer 7

find the unite of the slope. Base * height

s * m/s^2 = m/s

so the area indicates how much velocity changes at a constant rate over a certain unit of time

Answer 8

It's the change in velocity. This is a no-brainer.

Answer 9

average velocity