Question on an integral

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Fusillo

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Hello! I'd like to ask for clarifications on this exercise. Here's a formula to express acceleration that looks like this:
a = [MATH] \dfrac{pr\left(\sin\left(\frac{{\pi}\left(\frac{t}{s}-1\right)}{2}\right)+1\right)}{2m}+a_0[/MATH]for t (time) in [0, 2*s]

I calculated antiderivatives to find v (velocity) and x (position):
v = [MATH] \dfrac{pr\left(t-\frac{2s\cos\left(\frac{{\pi}\left(\frac{t}{s}-1\right)}{2}\right)}{{\pi}}\right)}{2m}+a_0t+v_0 [/MATH]x = [MATH] \dfrac{pr\left(\frac{t^2}{2}-\frac{4s^2\sin\left(\frac{{\pi}\left(\frac{t}{s}-1\right)}{2}\right)}{{\pi}^2}\right)}{2m}+\dfrac{a_0t^2}{2}+v_0t+x_0 [/MATH]
Now assume a0, v0, x0 = 0.
When replacing t=0, acceleration and velocity are zero, but x isn't. Why?
 
Fusillo said:
Hello! I'd like to ask for clarifications on this exercise. Here's a formula to express acceleration that looks like this:
a = [MATH] \dfrac{pr\left(\sin\left(\frac{{\pi}\left(\frac{t}{s}-1\right)}{2}\right)+1\right)}{2m}+a_0[/MATH]for t (time) in [0, 2*s]

I calculated antiderivatives to find v (velocity) and x (position):
v = [MATH] \dfrac{pr\left(t-\frac{2s\cos\left(\frac{{\pi}\left(\frac{t}{s}-1\right)}{2}\right)}{{\pi}}\right)}{2m}+a_0t+v_0 [/MATH]x = [MATH] \dfrac{pr\left(\frac{t^2}{2}-\frac{4s^2\sin\left(\frac{{\pi}\left(\frac{t}{s}-1\right)}{2}\right)}{{\pi}^2}\right)}{2m}+\dfrac{a_0t^2}{2}+v_0t+x_0 [/MATH]
Now assume a0, v0, x0 = 0.
When replacing t=0, acceleration and velocity are zero, but x isn't. Why?
Click to expand...
Please show us - how "you" concluded that at t=0 - acceleration and velocity are zero.

Also post the whole assignment - EXACTLY verbatim.
 
go back to the integration of the velocity and let [MATH]t=0[/MATH]
[MATH]x(0) =\dfrac{2 pr s^2}{m \pi^2} + x_0\\ \text{now if you want $x(0)=0$ then $x_0 = -\dfrac{2 pr s^2}{m \pi^2}$}[/MATH]
It's easier to see if you name the constant of integration [MATH]C[/MATH] instead of [MATH]x_0[/MATH]
 
Subhotosh Khan said:
Please show us - how "you" concluded that at t=0 - acceleration and velocity are zero.

Also post the whole assignment - EXACTLY verbatim.
Click to expand...
Hi, thank you for the answer. It's not an exercise or homework, I'm trying to program an autopilot for a vehicle and that's the acceleration formula for the engine.
a0, v0, x0 are the initial acceleration, velocity, position, and when I start the car (t=0) they're all zero.
p, r, m, s are constants.
To conclude that acceleration and velocity are zero you just need to replace t in the formulas I provided.
In the acceleration formula it is: (sin(-pi/2) - 1) which makes the numerator go to zero.
In the velocity formula also the numerator is zero since cos(-pi/2) is zero.
 
Romsek said:
go back to the integration of the velocity and let [MATH]t=0[/MATH]
[MATH]x(0) =\dfrac{2 pr s^2}{m \pi^2} + x_0\\ \text{now if you want $x(0)=0$ then $x_0 = -\dfrac{2 pr s^2}{m \pi^2}$}[/MATH]
It's easier to see if you name the constant of integration [MATH]C[/MATH] instead of [MATH]x_0[/MATH]
Click to expand...
The vehicle in t=0 hasn't started moving yet, so it should be x = x0 = 0
 
I see what the difficulty is.

We haven't been rigorous enough.

Given [MATH]a(t)[/MATH], and assuming time starts at [MATH]t=0[/MATH]
[MATH]v(t) = \displaystyle \int \limits_0^t ~a(\tau)~d\tau + v_0[/MATH]
and likewise

[MATH]x(t) = \displaystyle \int \limits_0^t ~v(\tau)~d\tau[/MATH]
Clipboard01.jpg

Plug all your zeros into those and see what you get.
 
Romsek said:
I see what the difficulty is.

We haven't been rigorous enough.

Given [MATH]a(t)[/MATH], and assuming time starts at [MATH]t=0[/MATH]
[MATH]v(t) = \displaystyle \int \limits_0^t ~a(\tau)~d\tau + v_0[/MATH]
and likewise

[MATH]x(t) = \displaystyle \int \limits_0^t ~v(\tau)~d\tau[/MATH]
View attachment 23934

Plug all your zeros into those and see what you get.
Click to expand...
a(0) = 0
v(0) = 0
x(0) = 0
That actually works, so my integral for x was incorrect after all? I used https://www.integral-calculator.com/. Thank you for the help I'll try to use that and see how it goes
 
Romsek said:
The form was correct, but it had to be evaluated at the limits like a definite integral.
Click to expand...
So in practice I had to calculate x(t) - x(0) with t=0, so that the "extra" gap would eliminate itself? In the picture you can see your version of x(t) in red, mine in magenta.
Cattura.PNG
The red function is your x(t), while the magenta one is mine.
 
You'll find that gap is equal to the quantity I mention in post #3.
 
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