Answered: An object is at the origin at time t=2…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

An object is at the origin at time t=2 , with velocity<1,2,3> at that time. The acceleration changes over time, and is given by the function < t², sin(t), t >. Find the path of the object.

Expert Solution

Step 1

The acceleration of the particle is given by the function

$a (t) = <t^{2}, \sin t, t> \frac{d v (t)}{d t} = <t^{2}, \sin t, t>$

Integrate both sides.

$v (t) = <\frac{1}{3} t^{3}, - \cos t, \frac{1}{2} t^{2}> + <a, b, c>$

Now, v(2)=<1,2,3>

Therefore,

$<1, 2, 3> = <\frac{8}{3}, - \cos 2, 2> + <a, b, c> <a, b, c> = <- \frac{5}{3}, 2 + \cos 2, 1>$

Thus,

$v (t) = <\frac{1}{3} t^{3}, - \cos t, \frac{1}{2} t^{2}> + <- \frac{5}{3}, 2 + \cos 2, 1> v (t) = <\frac{1}{3} t^{3} - \frac{5}{3}, - \cos t + (2 + \cos 2), \frac{1}{2} t^{2} + 1>$

Step 2

Now,

$\frac{d s (t)}{d t} = <\frac{1}{3} t^{3} - \frac{5}{3}, - \cos t + (2 + \cos 2), \frac{1}{2} t^{2} + 1> s (t) = <\frac{1}{12} t^{4} - \frac{5}{3} t, - \sin t + (2 + \cos 2) t, \frac{1}{6} t^{3} + t> + <d, e, f>$

Now, s(2)=<0,0,0>

$s (t) = <\frac{1}{12} t^{4} - \frac{5}{3} t, - \sin t + (2 + \cos 2) t, \frac{1}{6} t^{3} + t> + <d, e, f> <0, 0, 0> = <\frac{4}{3} - \frac{10}{3}, - \sin 2 + (2 + \cos 2) 2, \frac{4}{3} + 2> + <d, e, f> <d, e, f> = <- 2, 4 - \sin 2 + 2 \cos 2, \frac{10}{3}>$

Thus, position of particle is

$s (t) = <\frac{1}{12} t^{4} - \frac{5}{3} t, - \sin t + (2 + \cos 2) t, \frac{1}{6} t^{3} + t> + <- 2, 4 - \sin 2 + 2 \cos 2, \frac{10}{3}> s (t) = <\frac{1}{12} t^{4} - \frac{5}{3} t - 2, - \sin t + (2 + \cos 2) t + 4 - \sin 2 + 2 \cos 2, \frac{1}{6} t^{3} + t + \frac{10}{3}>$

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