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Name: The horizontal coordinates of a Frisbee in a strong wind are given by
Uploaded: 2024-03-20T06:57:47.000Z
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Description: The horizontal coordinates of a Frisbee in a strong wind are given by x = -12t + 4t^2 and y = 10t - 3t^2, where x and y are in meters, and t is in seconds.(a) What is the acceleration of the Frisbee? Give a magnitude and a direction,measuring angles

Principles of Physics: A Calculus-Based Text

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ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter3: Motion In Two Dimensions

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Problem 1P: A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.00...

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The horizontal coordinates of a Frisbee in a strong wind are given by x = -12t + 4t^2 and y = 10t - 3t^2, where x and y are in meters, and t is in seconds.(a) What is the acceleration of the Frisbee? Give a magnitude and a direction,measuring angles from the positive x direction.(b) What is the magnitude of the velocity at t = 2.0 s, accurate to the nearest m/s?

Expert Solution

Step 1

Given,

Horizontal (x , y) coordinate of frisbee is

$x = - 12 t + 4 t^{2} y = 10 t - 3 t^{2}$

(a) the acceleration of the Frisbee

We know that the acceleration is

$a_{Frisbee} = \frac{d^{2} x}{{dt}^{2}} + \frac{d^{2} y}{{dt}^{2}}$

Now acceleration a_x

_{$\frac{dx}{dt} = \frac{d}{dt} (- 12 t + 4 t^{2}) \frac{dx}{dt} = - 12 + 8 t again differentiation \frac{d^{2} x}{{dt}^{2}} = \frac{d}{dt} - 12 + 8 t \frac{d^{2} x}{{dt}^{2}} = 8 dx = 8 m / s^{2}$}

Now, acceleration a_y

_{$\frac{dy}{dt} = \frac{d}{dt} (10 t - 3 t^{2}) \frac{dy}{dt} = 10 - 6 t again differentiation \frac{d^{2} y}{{dt}^{2}} = \frac{d}{dt} (10 - 6 t) \frac{d^{2} y}{{dt}^{2}} = - 6 dy = - 6 m / s^{2}$}

So, the acceleration of Frisbee is

$\vec{a} = 8 \overset{\land}{i} - 6 \overset{\land}{j} m / s^{2}$

Now the magnitude is given as

$a = \sqrt{8^{2} + {(- 6)}^{2}} a = \sqrt{100} a = 10 m / s^{2}$

Now the direction is

$\tan θ = \frac{6}{8} θ = \tan^{- 1} (\frac{6}{8}) θ = 36 . 87^{\circ}$

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