AP Calculus AB Free Response Question

[scottshannon]

I needed help on this AB Calculus Free Response Question:

A particle moves along the x-axis do that its velocity $v$ at time $t \ge 0$ is given by $v(t) = \sin( t^2)$ for the interval $0 \le t \le \sqrt{5\pi}$. The position of the particle at time $t$ is $x(t)$ and its position at time $t = 0$ is $x(0) = 5$ .

a) Find the acceleration of the particle at time $t = 3$.
b) Find the total distance traveled by the particle from time $t = 0$ to $t = 3$.
c) Find the position of the particle at time $t =3$.
d) For $0 \le t \le \sqrt{5\pi}$, find the time $t$ at which the particle is farthest to the right. Explain your answer.

I believe that I can do a) but I am not sure about the rest.

 

[Ackbach]

For part b, can you think about a particle changing direction? When would that occur? If a particle goes only in one direction, then what's the relationship between displacement and total distance traveled? If a particle changes direction, what's the relationship between displacement and total distance traveled?

For part d, is there some aspect of differential calculus you can think of that you could bring to bear on find the biggest value of the position function?

 

[scottshannon]

I needed help on this AB Calculus Free Response Question:

A particle moves along the x-axis do that its velocity $v$ at time $t \ge 0$ is given by $v(t) = \sin( t^2)$ for the interval $0 \le t \le \sqrt{5\pi}$. The position of the particle at time $t$ is $x(t)$ and its position at time $t = 0$ is $x(0) = 5$ .

a) Find the acceleration of the particle at time $t = 3$.
b) Find the total distance traveled by the particle from time $t = 0$ to $t = 3$.
c) Find the position of the particle at time $t =3$.
d) For $0 \le t \le \sqrt{5\pi}$, find the time $t$ at which the particle is farthest to the right. Explain your answer.

I believe that I can do a) but I am not sure about the rest.

My problem is that I don't know how to figure displacement unless I can integrate. If you could tell me how to work the problem rather than give me a question, I would be most appreciative.

 

[scottshannon]

So between $0$ and $\sqrt{\pi}$ the velocity is positive. Between $\sqrt{\pi}$ and $\sqrt{2\pi}$ the velocity is negative. How does that help me if I cannot integrate $\sin\left({t^2}\right)$?

 

[Euge]

I needed help on this AB Calculus Free Response Question:

A particle moves along the x-axis do that its velocity $v$ at time $t \ge 0$ is given by $v(t) = \sin( t^2)$ for the interval $0 \le t \le \sqrt{5\pi}$. The position of the particle at time $t$ is $x(t)$ and its position at time $t = 0$ is $x(0) = 5$ .

a) Find the acceleration of the particle at time $t = 3$.
b) Find the total distance traveled by the particle from time $t = 0$ to $t = 3$.
c) Find the position of the particle at time $t =3$.
d) For $0 \le t \le \sqrt{5\pi}$, find the time $t$ at which the particle is farthest to the right. Explain your answer.

I believe that I can do a) but I am not sure about the rest.

Hi scottshannon,

For b), the total distance traveled from $t = 0$ to $t = 3$ is $\int_0^3 v(t)\, dt$. You have $v(t)$, but note that you cannot simplify the integral any further.

For c), use the fundamental theorem of calculus to get $x(3) = x(0) + \int_0^3 v(t)\, dt = 5 + \int_0^3 \sin(t^2)\, dt$.

Finally for d), use the first (or second) derivative test to find the time $t$ that maximizes $x(t)$.

 

[Evgeny.Makarov]

For b), the total distance traveled from $t = 0$ to $t = 3$ is $\int_0^3 v(t)\, dt$.

If we believe in the distinction between distance and displacement as on this figure from Wikipedia,

View attachment 3984

then the total distance is $\int_0^3 \lvert v(t)\rvert\, dt$.

 

[Euge]

If we believe in the distinction between distance and displacement as on this figure from Wikipedia,
then the total distance is $\int_0^3 \lvert v(t)\rvert\, dt$.

Yes, I should have added that comment. Thanks for adding that point. I think the picture is slightly misleading though. It doesn't quite illustrate that in one-dimensional motion, the displacement under a given time interval is signed area of the velocity-time graph over that interval, whereas the distance is the unsigned area of the velocity-time graph over that interval.

 

[scottshannon]

Thank you...I understand the need for absolute value so as to calculate the total distance, so that the be negative will not subtract from the positive displacement.

Short of writing the infinite series for $\sin\left(t^2\right)$ and then integrating term by term, I don't know if there is a way to come up with a numerical value for . If $f(t)=\sin\left(t^2\right)$ then

\(\displaystyle F(t) = \int_{t}^{0} \,\sin({t}^{2}) +5\)

I am not sure that I am using latex right. Is my latex being translated to symbols? If not can you help me use latex?

 

[scottshannon]

On part d I am not sure how to take the first derivative.

 

[scottshannon]

Should the integral be written $\int_{0}^{{t}^{2}} \,sin(u)$ or
$\int_{0}^{t} \,sin({u}^{2}$

 

[MarkFL]

I am thinking you need:

\(\displaystyle x(t)=\frac{1}{2t}\int_0^{t^2}\sin(u)\,du\)

Do you see why?

 

[Euge]

Hello again scottshannon,

The derivative of $x(t)$ is $v(t)$, which is already given. You won't need to worry about finding an approximate value for $\int_0^3 |\sin t^2|\, dt$. However, if you need it, the value is approximately $1.702$.

 

[scottshannon]

I don't see why.. I am sorry

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I don't think that having the integrand sin(u) with limits t^2 and 0 is the same as sin(u^2) with limits t and 0 (I am too stupid to use latex sorry)

 

[scottshannon]

how do you get the cursive letters that you use for x(t) and v(t)..I can't do that

 

[scottshannon]

$\int_{0}^{u} \,\sin\left({{t}^{2}}\right)dt$

 

[Euge]

how do you get the cursive letters that you use for x(t) and v(t)..I can't do that

If you put v(t) in between two dollars signs, you get $v(t)$. Similarly for $x(t)$. Try it and see what you get.

 

[scottshannon]

The function is $v(t) = \sin\left({{t}^{2}}\right)$ , so it seems that the integral

should be $\int_{0}^{u} \,\sin\left({{t}^{2}}\right)dt$ not

$\int_{0}^{{t}^{2}} \,\sin\left({u}\right)du$

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Perhaps I do not understand the fundamental theorem of calculus well enough

 

[MarkFL]

I purposefully gave you an integral expression for $x(t)$ so that when you tried maximizing by differentiating, you would realize that you already had the derivative of $x(t)$ in the given $v(t)$. :D

 

[Evgeny.Makarov]

I am thinking you need:

\(\displaystyle x(t)=\frac{1}{2t}\int_0^{t^2}\sin(u)\,du\)

Indeed, \(\displaystyle \frac{d}{dt}\int_0^{t^2}\sin(u)\,du=2t\sin(t^2)\), but \(\displaystyle \frac{d}{dt}\left(\frac{1}{2t}\int_0^{t^2}\sin(u)\,du\right)\ne \sin(t^2)\).

 

[MarkFL]

Indeed, \(\displaystyle \frac{d}{dt}\int_0^{t^2}\sin(u)\,du=2t\sin(t^2)\), but \(\displaystyle \frac{d}{dt}\left(\frac{1}{2t}\int_0^{t^2}\sin(u)\,du\right)\ne \sin(t^2)\).

You mean we can't just ignore the product rule this one time?(Wasntme)