Confused about integrating acceleration to get velocity

[late347]

Homework Statement

acceleratin as function of time
$a(t)= 2t+1$
we know that v(0)=0
and s(0)=0

find v(t)
find v(5)
find s(t)
find s(3)
and I was thinking about also what happens when t is negative number,
is it possible to find also v(-2)?
what about s(-3)?

Homework Equations

integration rules, I suppose

The Attempt at a Solution

I did the problem by
antidifferentiating the acceleration function
$a(t) = v'(t)$
$\int_{}^{} a(t) dt = v(t)$
$v(t) = t^2 +t +c_{1}$
it was given that v(0)=0
now we can plug in v=0 and deduce that $c_{1}$ =0
v(5)=30for the distance as function of time s(t)
$s(t)= \int_{}^{} v(t) dt$
$s(t) =0.333...*t^3+ 0.5*t^2 +c_{2}$
it was given that s(0) =0
hence deduce that $c_{2}=0$
s(3)=13.5 but I was little bit confused when my teacher's solution showed simply that
$v(t)= \int_{0}^{t} a(t)dt$

I was confused about three things essentially

1.)why the lowerbound for definite integral has to be 0 as it was in my teachers solution

2.)why the upperbound is variable t, it seems like there's so many things happening in my teachers solution so I got a bit confused. I think the integral function notation is much clearer in my opinion.

3.)and what happens if I want to calculate v(-2), because... why not? what will be the upperbound and nthe lowerbound if I want to use my teacher's technique of definite integral?

OK I Know intuitively that when you have acceleration as function of time,
then the velocity has to be the integral of some sort ( dt is the width of a small rectangle, and multiply it by the a(t) the height to get the thin rectangle area, then you do the Riemann sum, velocity is the summed area)

It seems as though I somehow got the correct and same answer as my teacher but I'm still little bit confused about the whole thing as I described above.

 

[Charles Link]

When they specify an initial condition at $t=0$, the formulas usually work for $t>0$, but most often don't work for $t<0$. e.g. if you throw a ball up at time $t=0$ at velocity $v_o$, and it responds to gravity, you don't have any information about the ball before the time $t=0$. You can write $s=v_ot-(1/2)gt^2$ for time $t>0$, but it doesn't apply for $t<0$.

 

[late347]

When they specify an initial condition at $t=0$, the formulas usually work for $t>0$, but most often don't work for $t<0$. e.g. if you throw a ball up at time $t=0$ at velocity $v_o$, and it responds to gravity, you don't have any information about the ball before the time $t=0$. You can write $s=v_ot-(1/2)gt^2$ for time $t>0$, but it doesn't apply for $t<0$.

ok, but mathematically speaking the integral function $v(t)= t^2+t$is defined in the real numbers, I suppose it makes more sense to think about the thing from the physics point of view though, like you suggested

the function itself does go under the t-axis such as v(-0.5)=-0.25

 

[Charles Link]

If the initial condition was made by a speedometer/velocity meter that measured the ball at t=0 even though it was thrown at t=-4 seconds, then it would make sense to use formulas to find $v(-3)$. In most cases, that is not the case.

 

[Ray Vickson]

Homework Statement

acceleratin as function of time
$a(t)= 2t+1$
we know that v(0)=0
and s(0)=0

find v(t)
find v(5)
find s(t)
find s(3)
and I was thinking about also what happens when t is negative number,
is it possible to find also v(-2)?
what about s(-3)?

Homework Equations

integration rules, I suppose

The Attempt at a Solution

I did the problem by
antidifferentiating the acceleration function
$a(t) = v'(t)$
$\int_{}^{} a(t) dt = v(t)$
$v(t) = t^2 +t +c_{1}$
it was given that v(0)=0
now we can plug in v=0 and deduce that $c_{1}$ =0
v(5)=30for the distance as function of time s(t)
$s(t)= \int_{}^{} v(t) dt$
$s(t) =0.333...*t^3+ 0.5*t^2 +c_{2}$
it was given that s(0) =0
hence deduce that $c_{2}=0$
s(3)=13.5but I was little bit confused when my teacher's solution showed simply that
##v(t)= \int_{0}^{t} a(t)dt #hna

I was confused about three things essentially

1.)why the lowerbound for definite integral has to be 0 as it was in my teachers solution

2.)why the upperbound is variable t, it seems like there's so many things happening in my teachers solution so I got a bit confused. I think the integral function notation is much clearer in my opinion.

3.)and what happens if I want to calculate v(-2), because... why not? what will be the upperbound and nthe lowerbound if I want to use my teacher's technique of definite integral?

OK I Know intuitively that when you have acceleration as function of time,
then the velocity has to be the integral of some sort ( dt is the width of a small rectangle, and multiply it by the a(t) the height to get the thin rectangle area, then you do the Riemann sum, velocity is the summed area)

It seems as though I somehow got the correct and same answer as my teacher but I'm still little bit confused about the whole thing as I described above.

The solution of $v^{\prime}(t) = a(t)$ with $v(0) = c$ is $v(t) = c + \int_0^t a(\tau) \, d\tau$. When $t=0$ the right-had-side is just $c + \int_0^0 \cdots = c$, and the derivative of the right-hand-side is just $a(t)$.

By the way: your teacher should NOT have used the notation $v(t) = \int_0^t a(t) \, dt,$ because that is making the same symbol "$t$" serve two very different roles in the same expression. Of course, an indefinite-integral such as $v(t) = \int a(t) \, dt + C$ is normal usage, but when you have a definite integral you should avoid it. That is why I used a different integration dummy variable $\tau$ in the formula I wrote. Anything other that $t$ itself would be perfectly OK, so we could have used $s$ or $t'$ or $w$ or any other among thousands of options.

When you have initial conditions specified at $t=0$ it is most convenient to let the lower limit of integration be 0. If you had initial conditions at $t=1$ it would be most convenient to have $\int_1^t \cdots$. Convenience is not the same thing as necessity, so if you want to do extra work, go ahead and feel free to choose a different lower limit, but just be sure to include an additive constant in your final formula.

 

[LCKurtz]

Homework Statement

acceleratin as function of time
$a(t)= 2t+1$
we know that v(0)=0
and s(0)=0
<snip>

but I was little bit confused when my teacher's solution showed simply that
$v(t)= \int_{0}^{t} a(t)dt$

I was confused about three things essentially

1.)why the lowerbound for definite integral has to be 0 as it was in my teachers solution

2.)why the upperbound is variable t, it seems like there's so many things happening in my teachers solution so I got a bit confused. I think the integral function notation is much clearer in my opinion.

To add to what others have said, since $v(t)$ is an antiderivative of $a(t)$, the fundamental theorem of calculus states that $v(c) - v(b) = \int_b^c a(t)~dt$. Since you know $v(0)= 0$ and you want $v(t)$, if you just put $b=0$ and $c=t$ you get $v(t)-0=\int_0^t a(t)~dt$. Remember that $t$ inside the definite integral is a dummy variable. Also note that if you had been given $v(5) =4$ as an initial condition instead of $v(0)=0$, you might have wanted to choose the integral from $5$ to $t$. It just saves a bit of writing, but there is no real problem with just working with indefinite integrals and constants of integration as you did.

[Edit] I see that Ray and others were busy posting while I was typing...

 

[fresh_42]

... but I was little bit confused when my teacher's solution showed simply that
$v(t)= \int_{0}^{t} a(t)dt$

I was confused about three things essentially

1.)why the lowerbound for definite integral has to be 0 as it was in my teachers solution

I think it would have been better to have written $v(t)= \int_{0}^{t} a(s)ds$ and not to use the same letter in two different meanings. Then we have that $v(t)$ is the anti-derivative of $a(t)$ which means by the fundamental theorem of calculus, that
$$
\int_{s=t_1}^{s=t_2} a(s) ds = v(t_2) - v(t_1)
$$
and with the settings $t_1=0$ and $t_2=t$ together with $v(0)=0$ you get the desired result. In general we can also write
$$
\int_{s=t_1}^{s=t_2} a(s) ds = \int_{s=0}^{s=t_2} a(s) ds - \int_{s=0}^{s=t_1} a(s) ds = [v(t_2) - v(0) ] - [v(t_1) - v(0)] = v(t_2) - v(t_1)
$$
and the constant term can be of any value and we could always integrate from $0$ on.

2.)why the upperbound is variable t, it seems like there's so many things happening in my teachers solution so I got a bit confused. I think the integral function notation is much clearer in my opinion.

see above

3.)and what happens if I want to calculate v(-2), because... why not? what will be the upperbound and nthe lowerbound if I want to use my teacher's technique of definite integral?

Same as what you did, with the hidden assumption that the equation for acceleration is the same in the past, which means that from some time on, breaking in the past turned into acceleration in the future.

 

[late347]

I think it would have been better to have written $v(t)= \int_{0}^{t} a(s)ds$ and not to use the same letter in two different meanings. Then we have that $v(t)$ is the anti-derivative of $a(t)$ which means by the fundamental theorem of calculus, that
$$
\int_{s=t_1}^{s=t_2} a(s) ds = v(t_2) - v(t_1)
$$
and with the settings $t_1=0$ and $t_2=t$ together with $v(0)=0$ you get the desired result. In general we can also write
$$
\int_{s=t_1}^{s=t_2} a(s) ds = \int_{s=0}^{s=t_2} a(s) ds - \int_{s=0}^{s=t_1} a(s) ds = [v(t_2) - v(0) ] - [v(t_1) - v(0)] = v(t_2) - v(t_1)
$$
and the constant term can be of any value and we could always integrate from $0$ on.

see above

Same as what you did, with the hidden assumption that the equation for acceleration is the same in the past, which means that from some time on, breaking in the past turned into acceleration in the future.

So, when you have the so-called dummy variable... I don't get it why it is called dummy variable? I thought that the variable matters because the definite integral should be like, essentially... the limit of the Riemann sum. So.. I suppose if we don't do the actual limit, but just think about the Riemann sum.

So if you have the $v(t)= \int_{0}^{t} a(s)ds$

It looks like you have something like acceleration in the vertical axis as the function value then you have the inputs in the horizontal s- axis. And then it looks like you have the differential ds, an infinitesimal horizontal width. Then you multiply it with the a(s) and get the thin rectangle. Then you sum the rectangles up to the point s=t. Because you do the sum from s=0 up to the t value on the s-axis. To my mind it looks like the lower bound is just fixed at 0 and the t can be moved on the s-axis to get different velocity.Obviosly the area "under the curve" is the same regardless of what we label the variable or the horizontal and vertical axes...? As long as the acceleration graph is the same graph I suppose...

So is this wrong thinking or a good idea? BTW I just had a math exam about this, I don't know how good I did, but I did manage to do a tough integration by parts problem! :)

 

[Charles Link]

It is good practice to use a dummy variable and write it as $v(t)=\int\limits_{0}^{t} a(t') \, dt'$. One reason for this is that occasionally you will take derivatives of expressions like this, e.g. $\frac{dv(t)}{dt}=a(t)$. It is helpful to not have the extra $t$ floating around, and to have a dummy variable when you do such an operation. $\\$ Editing: Perhaps $s$ is also not a good choice for the dummy variable, because $s$ is often used for distance, so I put a $t'$ in there instead.

 

[fresh_42]

So, when you have the so-called dummy variable... I don't get it why it is called dummy variable?

Sorry, me neither. I don't use this term in this context. The only "´dummy" I see is $ds$ to distinguish the variable $s$ from constants like the the integration boundaries $0,t$ or the $2$ as given in the function of $a(s)$.

I thought that the variable matters because the definite integral should be like, essentially... the limit of the Riemann sum. So.. I suppose if we don't do the actual limit, but just think about the Riemann sum.

So if you have the $v(t)= \int_{0}^{t} a(s)ds$

It looks like you have something like acceleration in the vertical axis as the function value then you have the inputs in the horizontal s- axis. And then it looks like you have the differential ds, an infinitesimal horizontal width. Then you multiply it with the a(s) and get the thin rectangle. Then you sum the rectangles up to the point s=t. Because you do the sum from s=0 up to the t value on the s-axis. To my mind it looks like the lower bound is just fixed at 0 and the t can be moved on the s-axis to get different velocity.

Yes, this is correct. You measure (sum up) all the gains (or losses) by acceleration from the time stamp $0$ to the "actual or current" time $t$ in order to get the accumulated velocity at $t$ (and add the velocity at the start $v(0)$ which is $0$ in your case).

Obviosly ...

Just as a side note: Be more careful with this word. I remember that I've once spent three days on a search for the trick in a calculation behind "obvious". Or was it "clearly" or "trivial"?

... the area "under the curve" is the same regardless of what we label the variable or the horizontal and vertical axes...? As long as the acceleration graph is the same graph I suppose...

Yes. I just took $s$ as the nearest choice to the given and within $\int_0^t a(s)ds$ constant time $t$, because I wanted to distinguish the variable of the function $a(s)$ from the integration boundary $s=t$. The constant $t$ becomes a variable only if we let it flow, i.e. if we write $t \mapsto v(t) = \int_0^t a(s)ds$. And of course you can name them whatever you want. However, people are used to acceleration, velocity and time, and e.g. $\Omega(\zeta) = \int_0^\xi o(\alpha) d\alpha$ would be problematic for several reasons:

• You would have to precisely define those functions $\Omega$ and $o(.)\,.$
• Lower case Greek letters like $\alpha$ are commonly used to name angles.
• $\Omega$ has already its own meaning.
• $o(.)$ has also a certain meaning.
• Whereas $v(t) = \int_0^t a(s)ds$ is recognized by everybody without further explanations, $\Omega(\zeta) = \int_0^\xi o(\alpha) d\alpha$ hints to a completely different set-up which isn't even physics.

So is this wrong thinking ...

It's not wrong, but a bit too close to the definition of an integral to be used on a daily basis. If you use integrals a lot, you won't always go down to Riemann sums, rather just integrate. But it is necessary to know the conception behind. It will be the fallback scenario if other methods don't work, e.g. when integration has to be done by numeric methods instead of closed expressions.

... or a good idea? BTW I just had a math exam about this, I don't know how good I did, but I did manage to do a tough integration by parts problem! :)

Depends on what you mean by wrong. As far as renaming the functions and variables is concerned, it is at least questionable, as I've explained. With respect to the (hidden) summation process of Riemann sums, it is the right understanding.

 

[Ray Vickson]

So, when you have the so-called dummy variable... I don't get it why it is called dummy variable? I thought that the variable matters because the definite integral should be like, essentially... the limit of the Riemann sum. So.. I suppose if we don't do the actual limit, but just think about the Riemann sum.

So if you have the $v(t)= \int_{0}^{t} a(s)ds$

It looks like you have something like acceleration in the vertical axis as the function value then you have the inputs in the horizontal s- axis. And then it looks like you have the differential ds, an infinitesimal horizontal width. Then you multiply it with the a(s) and get the thin rectangle. Then you sum the rectangles up to the point s=t. Because you do the sum from s=0 up to the t value on the s-axis. To my mind it looks like the lower bound is just fixed at 0 and the t can be moved on the s-axis to get different velocity.Obviosly the area "under the curve" is the same regardless of what we label the variable or the horizontal and vertical axes...? As long as the acceleration graph is the same graph I suppose...

So is this wrong thinking or a good idea? BTW I just had a math exam about this, I don't know how good I did, but I did manage to do a tough integration by parts problem! :)

If you have ever done any computer coding, then you have probably met with the concept of global vs. local variables. For example, when writing a subroutine (or sub-program, or callable procedure), you may have variables inside the procedure that are created when you start the procedure and disappear when you leave the procedure. Those are variables that are "local" to the inner workings of the procedure; they are not delivered to the calling program. You could call them "dummy" variables, although computer people do not normally use that terminology. One thing you must never, ever do, is use the same variable name for a local variable as for the global variable that "calls" the procedure. For example, you may have a procedure called velocity(a,t), called by a function a(.) and a terminal time t. Its workings might be:

procedure velocity(a,t);
local x;
v =Integrate(a(x),x = 0 ..t);
return v;
end;

The program would not work if you said v = integrate(a(t),t=0..t); the computer would deliver an error message and grind to a halt.