Mathematics, 19.09.2019 08:00 juliashalnev

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion
s = 4 sin(πt) + 3 cos(πt),
where t is measured in seconds. (round your answers to two decimal places.)
a. find the average velocity during each time period.
(i) [1, 2]
(ii) [1, 1.1]
(iii) [1, 1.01]
(iv) [1, 1.001]
b. estimate the instantaneous velocity of the particle when t = 1.

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This looks like an exercise that's building toward the idea of a derivative.

These calculations are done best with a calculator, but here's how the first interval is used:

Average velocity = (position at 2 - position at 1) / (2 - 1) It's really distance divided by time!

Position at t = 2: $4\sin{(2\pi)}+3\cos{(2\pi)}=4(0)+3(1)=3$
Position at t = 1: $4\sin{(\pi)}+3\cos{(\pi)}=4(0)+3(-1)=-3$

So over the interval [1, 2] the average velocity is $\frac{3-(-3)}{2-1} = \frac{6}{1} = 6$

I used a spreadsheet to calculate the average velocity over the other intervals and a couple of shorter ones, too. (See attached image.)

As these intervals get shorter (the right endpoint is approaching 1), the average velocity gets closer and closer to the instantaneous velocity. An estimate would be -12.6.

The displacement (in centimeters) of a particle moving back and forth along a straight line is given

The displacement (in centimeters) of a particle moving back and forth along a straight line is given

Answer from: NehemiahPate2006

Average velocity is the change in the position of an object divided by the

time the object takes to make the change.

A. (i) The average velocity during the time period, [1, 2] is 6 cm/s

(ii) The average velocity during the time period, [1, 1.1] is approximately -10.892 cm/s.

(iii) The average velocity during the time period, [1, 1.01] is approximately -12.416 cm/s.

(iv) The average velocity during the time period, [1, 1.001] is approximately -12.566 cm/s.

B. The instantaneous velocity when t = 1 is approximately -12.566 cm/s.

Reasons:

A. In the cyclical motion of the particle, the average velocity depends on

the location.

$Average \ velocity = \dfrac{\Delta x}{\Delta t} = \overline v$

The given equation of motion is; s = 4·sin(π·t) + 3·cos(π·t)

From the graph of the equation, we have;

The period of the equation is 2 seconds

A peak occurs at t ≈ 0.295 seconds followed by a trough at 1.295

seconds.

Therefore;

Average velocity between points t = 1, and t = 2, [1, 2] will be positive and

average velocities between points t = 1 to points 1 < t < 1.295 will be

negative

(i) Period [1, 2]

$\overline v= \dfrac{(4 \cdot sin(\pi \times 2) + 3\cdot cos(\pi \times 2)) - (4 \cdot sin(\pi \times 1) + 3\cdot cos(\pi \times 1)) }{2 - 1} = 6$

The average velocity during the time period, [1, 2] is 6 cm/s.

(ii) Period [1, 1.1]

$\overline v= \dfrac{(4 \cdot sin(\pi \times 1.1) + 3\cdot cos(\pi \times 1.1)) - (4 \cdot sin(\pi \times 1) + 3\cdot cos(\pi \times 1)) }{1.1 - 1} \approx -10.892$

$\overline v_{1, \, 1.1}$ ≈ -10.892 cm/s

(iii) Period [1, 1.01]

$\overline v= \dfrac{(4 \cdot sin(\pi \times 1.01) + 3\cdot cos(\pi \times 1.01)) - (4 \cdot sin(\pi \times 1) + 3\cdot cos(\pi \times 1)) }{1.01 - 1} \approx -12.416$

$\overline v_{1, \, 1.01}$ ≈ -12.416 cm/s

(iv) Period [1, 1.001]

$\overline v= \dfrac{(4 \cdot sin(\pi \times 1.001) + 3\cdot cos(\pi \times 1.001)) - (4 \cdot sin(\pi \times 1) + 3\cdot cos(\pi \times 1)) }{1.001 - 1} \approx -12.552$

$\overline v_{1, \, 1.001}$ = -12.552 cm/s

B. The instantaneous velocity is the value of the average velocity as the a

time interval for the change in position approaches zero seconds.

$\lim \limits_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt} = \dfrac{d}{dt} ( {4 \cdot sin(\pi \cdot t) + 3\cdot cos(\pi \cdot t)) } = 4 \cdot cos(\pi \cdot t) - 3\cdot sin(\pi \cdot t)$

$\lim \limits_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt} = v(t) = 4 \cdot cos(\pi \cdot t) - 3\cdot sin(\pi \cdot t)$

Therefore, when t = 1, we have;

v(1) = 4·cos(π × 1) - 3·sin(π × 1) ≈ -12.566

The instantaneous velocity at t = 1, v(1) ≈ -12.566 cm/s

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Answer from: Quest

Free points baby. free points baby. free points baby. hell yeah!

Answer from: Quest

Graph a should be your answer

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