Mathematics, 25.12.2019 06:31, Jasten

Marla is running clockwise around a circular track. she runs at a constant speed of 4 meters per second. she takes 46 seconds to complete one lap of the track. from her starting point, it takes her 12 seconds to reach the northernmost point of the track. impose a coordinate system with units in meters, the center of the track at the origin, and the northernmost point on the positive y-axis. (round your answers to two decimal places.) (a) give marla's coordinates at her starting point.

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answered: thil54451

(-29.21,-1.99) , Marla's position at the starting point!

Step-by-step explanation:

Given the information, we can design the coordinate system as instructed in the question.

Center of the circle = (0,0)

Northern most point is on the y-axis = (0,r)

Finding r:

given the velocity (4 m/s) and the time she takes for one complete lap (46sec).

we can use the formula:

$v = r \omega$

where v is the velocity and $\omega$ is the angular velocity is (rad/sec)

the given time to complete the lap is given in rps (rev/sec) (it takes 46 sec to complete one full revolution i.e 1rev/46sec). we first need to convert this unit into rad/sec so that we can use it in the formula above.

Unit conversion:

$\omega = \dfrac{1\,\text{rev}}{46\,\text{sec}}\times\dfrac{2\pi\,\text{rad}}{\text{rev}}$

$\omega = \dfrac{\pi}{23} \approx 0.1366\,\text{rad/sec}$ , this tells you that marla moves 0.1366 radians (or 7.28 degrees) every sec.

Back to formula

$v = r \omega$

4 = r(0.1366)

$r = 29.284 \text{m}$ now we know that she's this much away from the centre. Hence:

the northern most post is = (0,r) = (0,29.28)

this is the point where she was at the time t = 12 sec.

To find her position at time t = 0 sec, we need to think in terms of radians (or angles), i.e. $\theta$

we can set $\theta = 0$ at the time where Marla was at the northern most point. this will be our reference point, and angles will be measured from here.

so at $\text{time}\,\,t = 12\,\, ,\theta = 0 \,\, ,\text{position} = (0,29.28)$

and since we know that Marla moves 0.1365 radians every second, and we also know that at the 12th second she was at 0 radians.

So we can work our way into know her position (in radians) at t=0 (or 12 seconds ago)

we can simply subtract 0.1366 from her northern most position (in radians) 12 times.

$\theta_0 = 0 - 12(0.1366)$

$\theta_0 = -1.639 \text{radians}$

alternatively this can also be done through integration:

using the equations of motion for angular motion.

$\omega = \dfrac{d\theta}{dt}$

$\int\limits^{t=0}_{t=12} {\omega} \, dt = \int\limits^{\theta_0}_{\theta=0}\, d\theta$

$\omega (0) - \omega (12) = \theta_0 - 0$

we know that omega is the constant angular velocity = 0.1366 rad/s

we can solve this equation to and find the same answer.

$0.1366(0) - 0.1366(12) = \theta_0 - 0$

$\theta_0 = -1.639$

Finally, the coordinate of the starting position:

We know the angle (measured from our reference point)

$\theta_0 = -1.639$ (for better visualization, convert it into degrees i.e -93.9 degrees)

knowing that this position lies at the 3rd quadrant. (-x,-y)

we should add 90 degrees, so that we are within one quadrant. (and now we are measuring our angles from the -x axis, instead of y-axis)

$\theta_0 = -1.639 +\pi/2$

$\theta_0 = -0.068$

and r

r = 29.284

we can use trigonometry to find the coordinates (if you've already assigned the signs of each coordinate, then you don't need to put the sign of the angles in formulas below)

$x = r \cos{(\theta)}$

$x = 29.284 \cos{(0.068)} = 29.21$

$y = r \sin{(\theta)}$

$y = 29.284 \sin{(0.068)} = 1.998 = 2$

knowing that we are in the 3rd quadrant (-x,-y):

(-x,-y) = (-29.21,-1.99), her position at the starting point!

answered: Guest

answer: b)

step-by-step explanation:

its a dot plot because since the score is not high numbers.

- r3ktforgood ☕

answered: Guest

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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