Area under v-t and a-t graphs

I know that the magnitude ( i.e. considering all area as positive) of area under v-t graph gives distance travelled by the particle in the given time. But I've been told that magnitude of area under a-t graph does not gives speed. Is this true? If it's true, then what does magnitude of area under a-t graph represents?

v-t = velocity (y-axis) vs time (x-axis)

a-t = acceleration (y-axis) vs time (x-axis)

• The area under an acceleration-time graph does give the (change in) speed, in exactly the same way that the area under a velocity-time graph gives the (change in) distance. In both cases the area is equal to the intergral of the acceleration/velocity wrt time. Jul 29, 2015 at 15:57
• @JohnRennie the only (trivial) subtlety is that $a=0$ forever does not mean $v=0$. At the same time, $v=0$ forever does not mean $x=0$, so this is not really a difference between the two. However, I suspect that high school textbooks typically gloss over this point in the latter, but not the former case.
– Danu
Jul 29, 2015 at 15:59
• @JohnRennie if your first statement is true then suppose a case when a ball is thrown vertically upward out of window at t=0 and area under a-t graph is considered till it again reaches the window from t=0. In this condn, area under a-t graph is not 0 but change in speed is 0 ( if we consider the throw in vaccum). So I now think that my statement is correct and your's wrong. But my second question yet remains... Jul 29, 2015 at 16:09
• @JohnRennie And area under v-t graph gives change in displacement or position. And the magnitude gives distance travelled in given time. So your sense in saying change in distance is wrong. But change in speed is right. Jul 29, 2015 at 16:14
• The graph of $a:t$ that you describe is a horizontal straight line i.e. constant acceleration, so the area under it will be $at$ giving us the equation $\Delta v = at$ or in the form it's more commonly taught to schoolchildren $v = u + at$. This correctly describes what happens to the thrown ball because the velocity changes from $+u$ when you throw the ball to $-u$ when it lands, giving a change of $2u$ not zero. Jul 29, 2015 at 16:18

What is $a(t)$? It is the instantaneous time rate "change" of velocity & not velocity at time $t$! What it does tell you is how fast the velocity changes in an infinitesimal time period centered at $t$ i.e. $t \underset{-}+ dt$.

You can write $\dot{v}(t) = \dfrac{dv}{dt} = a(t)$. Then you want to find what change has occured to the velocity after starting from $t_0$ to some variable time $t$.

What should you do?

Since, the acceleration is not constant, we cannot use $\Delta v = a\Delta t$. However, if we make $\Delta t$ infinitesimally small, acceleration can be considered as constant (as it has not enough time to get changed), then you can find the change in velocity during this infintesimal time-interval i.e. $dv = a(t)dt$. But you wanted to know the change in velocity after travelling a time-period $t - t_0$. So, divide this period into $n$ finite sub-intervals of length $\Delta t$ & find the change for each sub-interval & then sum it. That is $\Delta{v} = \sum_{i =1}^n a(t)\Delta t$ . But as said, since the acceleration is not constant, we have to make $\Delta t$ very, very infinitesimal; in order to do this you have to make $n$ larger & larger & this can be written as $$\lim_{n \to \infty} \sum_{i =1}^n a(t)\Delta t$$ which is what we know as $$\int_{t_0}^{t} a(t)dt$$. So, what does it measure?? It measures change in velocity & not velocity.

However, you can really find velocity at a certain time $t$ using this integral as $$v(t) = \int_{t_0}^{t} a(t)dt + v(t_0)$$.

$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}$Velocity is the derivative of $s$ with respect to time

$$v = \dv{s}t$$

Take the integral of both sides
Integral is the area under the curve

$$\int v(t) \text d t = s$$

Acceleration is the derivative of velocity with respect to time

$$a = \dv v t$$

Take the integral of both sides

So the the area under $a$ with respect to $t$ should be velocity

$$\int a(t) \text d t= \text{Velocity} =v$$

Assume $a$ is constant and velocity starts at zero then you get

$$a \cdot t = v$$

For the ball thrown in the air I think you need to look at from negative acceleration on the way up as the it is the wrong direction and positive acceleration on the way down canceling out. If you drew a graph of $v$ versus $t$ it would be an exact V with the bottom of the at 0. The slope (acceleration) of the left is negative.