Physics, 16.10.2019 03:30 sophiamoser

Atelevision camera is positioned 4000 ft from the base of a rocket launching pad. the angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. let's assume the rocket rises vertically and its speed is 900 ft/s when it has risen 3000 ft.

a. how fast is the distance from the television camera to the rocket changing at that moment?

b. if the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

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Answer from: adam1299

a ) 540 ft /s

b ) .144° /s

Explanation:

Let at any moment h be the height of the rocket . The distance of rocket from camera will be R

R ² = 4000² + h²

Differentiating both sides with respect to t

R dR/dt = h dh/dt

dR/dt = h/R dh/dt

a ) Given speed of rocket

dh/dt = 900

h = 3000

R² = 4000² + 3000²

R = 5000

dR/dt = h/R dh/dt

= (3000 / 5000 ) X 900

dR/dt = 540 ft /s

b ) Let θ be angle of elevation at the moment .

4000 / R = cosθ

Differentiating with respect to t

- 4000 x 1 / R² dR/dt = - sinθ dθ / dt

4000 x ( 1/ 5000² ) x 540 = 3 /5 x dθ / dt

.0864 = 3/5 dθ / dt

dθ / dt = .144° /s

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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