0
$\begingroup$

To give you an example, say you're paragliding and in this scenario, your velocity and height starts as:

$t_0 : V_{y-initial} = 0\,m/s,\quad V_{x-initial} = 15\,m/s, \quad h_{initial} = 180\,m$

From this state, you release a rock that you were holding. Now assume no air resistance for the rock, and as such, $V_x = 15\,m/s$ for its entire fall. $|V_y| = |gt|$ where we assume $g = 10\,m/s^2$.

When the rock has hit the ground, how far has it travelled?

  • Q1: Which functions am I seeking? Velocity or distance? (To calculate total distance travelled)

To find the time $t_f$ when the rock has hit the ground, we put:

$$S_y(t) = 180 \iff \int V_y(t) = 5t^2 = 180 \implies t = 6\,s$$

Length of curve, i.e. total distance travelled is then calculated as:

$$L = \int^6_0\sqrt{V_x(t)^2+V_y(t)^2} = \int^6_0\sqrt{15^2+(10t)^2} \approx 209\,m$$

  • Q2: Is this answer even reasonable? Here's how I think: It must fall at least $180$ meters, which it fulfills. However, during the $6$ seconds, the minimum distance travelled in $x$-axis is $15\cdot6 = 90\,m$. This means the distance travelled must be greater than $180+90\,m$. It does not fulfill this. (I'm not sure where I'm wrong here).

I can solve for $V_{tot}$ as it's a net velocity between $V_x$ and $V_y$. Pythagoras' gives us,

$$V_{tot} = \sqrt{V_x^2+V_y^2} = \sqrt{15^2+(10t)^2}$$

And as such, it's a function we're looking to find the length of:

$$L = \int^6_0\sqrt{1+\sqrt{15^2+(10t)^2}}$$

  • Q3: Why is this wrong?
$\endgroup$

1 Answer 1

3
$\begingroup$

A1: I do not really understand you question. You did it correctly in the next paragraph.

A2: Your argument for why the length is wrong is wrong. $209\text{ m}$ is correct. When you add the lengths you assume they are in the same direction, but they are perpendicular. This means the answer should be atleast and just a bit more than $\sqrt{180^2+90^2}\approx 201$ by using pythagoras.

A3: That formula is for non-parametric curves (I do not know the proper name). I you want to use that you have to convert it from parametric to non-parametric. \begin{align*} S_y&=180-5t^2\\ S_x&=15t \end{align*} We want to express $S_y$ as a function of $S_x$. This gives us \begin{align*} S_y&=180-5\left(\frac{S_x}{15}^2\right)\\ \frac{dS_y}{dS_x}&=-\frac 2{45}S_x \end{align*} Now we use the formula \begin{align*} L&=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)}\,dx\\ &=\int_0^{90}\sqrt{1+\left(-\frac 2{45}x\right)}\\ &\approx 209 \end{align*}

$\endgroup$
2
  • $\begingroup$ Nailed it, good work. $\endgroup$ Dec 8, 2014 at 11:10
  • $\begingroup$ Sorry for late response. You made everything clear. Thanks a lot :) $\endgroup$
    – B. Lee
    Dec 8, 2014 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.