Answered: Velocity and acceleration from position…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Velocity and acceleration from position Consider the following position function.
a. Find the velocity and speed of the object.
b. Find the acceleration of the object.

r(t) = ⟨3 sin t, 5 cos t, 4 sin t⟩, for 0 ≤ t ≤ 2π

Expert Solution

Step 1 Given information:

$r (t) = ⟨ 3 \sin t, 5 \cos t, 4 \sin t ⟩, for 0 \leq t \leq 2 π$ .

a. To find: The velocity and speed of the object.
b. To find: The acceleration of the object.

Step 2

The velocity of the object is given by :

$\begin{array}{rcl} v (t) & = & r' (t) \\ = & <(3 \sin t)', (5 \cos t)', (4 \sin t)'> \\ = & <3 \cos t, - 5 \sin t, 4 \cos t> \end{array}$

Velocity: $\begin{array}{rcl} v (t) & = & <3 \cos t, - 5 \sin t, 4 \cos t> \end{array}$

Step 3

Now,

$\begin{array}{rcl} Speed : s (t) & = & | r' (t) | \\ = & \sqrt{{(3 \cos t)}^{2} + {(- 5 \sin t)}^{2} + {(4 \cos t)}^{2}} \\ = & \sqrt{9 \cos^{2} t + 25 \sin^{2} t + 16 \cos^{2} t} \\ = & \sqrt{25 \cos^{2} t + 25 \sin^{2} t} \\ = & \sqrt{25 (\cos^{2} t + \sin^{2} t)} \\ = & 5 \sqrt{(\cos^{2} t + \sin^{2} t)} \\ = & 5 \sqrt{1} \\ = & 5 \end{array}$

Hence, the speed of the object is $s (t) = 5$ .

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