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3 years ago

11

A buoy floating in the ocean is bobbing in simple harmonic motion with period 5 seconds and amplitude 4ft. Its displacement d fr

om sea level at time =t0 seconds is −4ft, and initially it moves upward. (Note that upward is the positive direction.) Give the equation modeling the displacement d as a function of time t.

1 answer:

Veronika [31]3 years ago

5 0

Answer:

$d=-4Cos(\frac{2\pi}{5}t)$

Step-by-step explanation:

We are given that

Period =5 s

Amplitude=4 ft

Displacement d from sea level at time $t=0s=-4 ft$

We have to find the modelling equation displacement d as a function of time.

We know that

The general equation of sinusoidal function is given by

$y(t)=Acos(Bt-C)+D$

B= $\frac{2\pi}{period}=\frac{2\pi}{5}$

When t=0, y=d=-4 ft, D=0

Substitute the values then we get

$-4=4Cos(\frac{2\pi}{5}(0)-C)+0$

$-4=4Cos(-C)$

$Cos(-C)=-1$

We know that Cos(-x)=Cos x

$Cos C=-1$

$Cos C=Cos \pi$ ( $cos(\pi)=-1$ )

$C=\pi$

Substitute the values then, we get

$d=4Cos(\frac{2\pi}{5}t-\pi)+0$

$d=4Cos(-(\pi-\frac{2\pi}{5}t))$

$d=4Cos(\pi-\frac{2\pi}{5}t)$

$Cos(\pi-x)=-Cosx$

$d=-4Cos(\frac{2\pi}{5}t)$

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Now, according to the question;

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The correct question is-

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If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

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Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

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