Wikipedia:Reference desk/Archives/Mathematics/2007 February 9

= February 9 =

intercept


What does it mean "It has slope -2/3 and x-intercept of 3"?


 * I assume you are talking about a line y = ax + b. "It has slope −2/3" means that a = −2/3, and "it has x-intercept 3" means that y = 0 when x = 3, so that b = 2. Thus y = −2x/3 + 2. --Andreas Rejbrand 06:33, 9 February 2007 (UTC)


 * I'm adding a picture to help visualize.
 * -2/3 — This is the slope of the line. This means that on a Cartesian plane (x-y plane), that for every two points you move to the right, you must move three points down (2→ and 3↓). Or, for every two points you move left, you must move three points up (2← and 3↑).
 * x-intercept of 3 — This means the line crosses the y=0 line at 3. In other words, it crosses the horizontal line, the x-axis at 3. The point is (0,3).
 * − Twas Now ( talk • contribs • e-mail ) 03:09, 10 February 2007 (UTC)

Vectors
I sat around the other day thinking how I could apply vectors. Then I thought:

If I had a pistol and wanted to see how far the bullet went if I shot it, and my pistol could fire a bullet at 350 m/s, how far would it go?

Assumptions:

Air resistance is negligible. I am firing with my pistol parallel to the ground (or at right angles). The terrain I am shooting out to is FLAT surface. The acceleration due to gravity is 9.8 metres per second per second downwards. I am holding the pistol 1.6m above the ground.

Am I correct in saying it would travel 1.4km before hitting the ground 4 seconds after being fired?


 * No, you are not. If we represent a vector V by writing $$\mathbf{V} = \underline{\mathbf{e}}\begin{pmatrix} x \\ y \end{pmatrix}$$ where (x, y) are its coordinates, then we have that the acceleration $$\mathbf{a}(t) = \underline{\mathbf{e}}\begin{pmatrix} 0 \\ -g \end{pmatrix}$$ where g = 9.8 m/s2. By integration, we find that the velocity $$\mathbf{v}(t) = \underline{\mathbf{e}}\begin{pmatrix} v_0 \\ -g t \end{pmatrix}$$ where v0 = 350 m/s. Integrating once again, we find that the position $$\mathbf{r}(t) = \underline{\mathbf{e}}\begin{pmatrix} v_0 t \\ h - \frac{1}{2} g t^2 \end{pmatrix}$$ where h = 1.6 m. The bullet reaches the ground when ry = 0 m, i.e. when $$h - \frac{1}{2} g t^2 = 0 \text{ m}$$. Solving for t, we get t = 0.571 ... s. Thus, $$r_x = v_0 t \approx 200 \text{ m} $$. --Andreas Rejbrand 22:16, 9 February 2007 (UTC)


 * This smells like a covert homework problem, so I'll say little. The interesting vector part of this is that we can separate the motion of the bullet into two independent components, one horizontal and one vertical. (If the viscosity of the air caused drag, could we do this?) The vertical component is pure gravity, the same as if we dropped the bullet from the given height. The horizontal component is steady inertial motion.
 * Recently a golf ball was hit from low Earth orbit. That is fun to analyze. --KSmrqT 22:46, 9 February 2007 (UTC)

Here's my working:

The acceleration vector of the bullet is -9.8j, which means the velocity is -9.8tj + c At time t = 0, v = 350i ==> 350i = 0j + c ==> c = 350i ==> v = 350i + (-9.8t)j ==> the position vector is 350ti + (-4.9t^2)j + C At time t = 0, position = 1.6j ==> 1.6j = C ==> p = 350ti + (-4.9t^2)j + 1.6j ==> p = 350ti + (1.6 - 4.9t^2)j Bullet hits the ground at vertical component = 0: 4.9t^2 = 1.6 ==> 49t^2 = 16 ==> t^2 = 16 ==> t = 4 So the bullet hits the ground after 4 seconds. At time t = 4, vertical component:

350 * 4 = 1400m


 * Here's your error: 49t^2 = 16 ==> t^2 = 16
 * Can you see it? --Andreas Rejbrand 22:33, 9 February 2007 (UTC)

Yeah, kicking myself. To "KSmrq," it's not homework. I haven't done maths at school for two years. I guess you probably assumed so because the answer works out to be exactly 200m, which was purely coincidental. Also, I searched Google for the speed at which a standard 9mm pistol is fired. It told me 200-500m/s, so I took the mean. If I went out and performed this, with all those assumptions in mind, would it still be 4/7 of a second? —Preceding unsigned comment added by 138.217.32.202 (talk • contribs)
 * I'll say a little more then. (And please sign and timestamp your posts, even if not logged in with an account, by typing four tildes: " ~ ".)
 * Because the bullet is fired horizontally, its initial velocity has a zero vertical component. The acceleration of gravity depends on neither the position nor the motion of the bullet; it is constant, and purely vertical. With no gravity, the bullet would travel parallel to the flat ground forever. It hits the ground for one reason: gravity pulls it down. The time it takes to cover the vertical distance to the ground depends on its initial height, its initial vertical velocity (stated to be zero), and the acceleration due to gravity. While it falls, it travels horizontally. How far it travels horizontally depends on how fast it is moving, and on how long it takes to fall to the ground.
 * If we have a higher muzzle velocity, the horizontal distance increases. If we fire the pistol from higher above the ground, the horizontal distance increases. If we tilt the aim upward, we get opposing effects. Less of the muzzle velocity is horizontal, which tends to decrease the horizontal distance covered; however, some of the muzzle velocity becomes vertical, which tends to increase the time to impact because the bullet goes up before it goes down. The distance traveled varies continuously as we adjust the aim from horizontal to straight up, and so we can find an optimal angle to yield the greatest horizontal distance. You might like to try to find that angle; doing so will help you better understand these calculations without straining your nascent abilities. --KSmrqT 05:01, 10 February 2007 (UTC)


 * I agree. Finding the optimal angle is an excellent exercise. It requires some (basic) calculus, though. If interested, I have actually derived this angle in a document (mechanics) at my website (the URL may be found on my user page). --Andreas Rejbrand 11:51, 10 February 2007 (UTC)


 * I don't really understand what you mean, but from the equation $$h - \frac{1}{2} g t^2 = 0 \text{ m}$$ we realize that the time of the entire motion does not depend on the initial speed of the bullet. Nevertheless, the horizontal distance $$r_x = v_0 t$$ (of course) does depend on it. --Andreas Rejbrand 23:19, 9 February 2007 (UTC)