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kow [346]
3 years ago
15

The displacement of a particle on a vibrating string is given by the equation s(t)=11+1/4sin(14PIt), where s is measured in cent

imeters and in seconds. Find the velocity of the particle after t seconds.
Mathematics
1 answer:
Lelechka [254]3 years ago
6 0
<span>Find the first derivative of the given expression.

</span>s(t)=11+\frac{ 1 }{ 4}\sin(14{\pi}t) &#10;\\&#10;\\s^{'}(t)=v(t)=\frac{ 7 }{ 2 }{\pi}\cos(14{\pi}t)<span>
</span>
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In option a, w < 10

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3 years ago
Mr. Brady is using a coordinate plane to design a treasure hunt for his students. The hunt begins at the flagpole. The first clu
dmitriy555 [2]

Answer:

The figure is attached and the total distance is 1031 feet.

Step-by-step explanation:

The graph is indicated in the attached figure.

For calculation of distance consider following

Point Flagpole is (0,0)

Point Clue1 is (0,5)

Point Clue2 is (6,0)

Point Clue3 is (0,-5)

So the distance is calculated as follows

d_{T}=[d_{FP\ to\ Clue1}+d_{Clue1\ to\ Clue2}+d_{Clue2\ to\ Clue3}]*distance\ per\ unit\\d_{T}=[\sqrt{(Clue1_x-FP_x)^2+(Clue1_y-FP_y)^2}+\sqrt{(Clue2_x-Clue1_x)^2+(Clue2_y-Clue1_y)^2}+\sqrt{(Clue3_x-Clue2_x)^2+(Clue3_y-Clue2_y)^2}]**distance\ per\ unit\\Substituting the values

d_{T}=[\sqrt{(0-0)^2+(5-0)^2}+\sqrt{(6-0)^2+(0-5)^2}+\sqrt{(0-6)^2+(-5-0)^2}]*50 \text{ feet}\\d_{T}=[\sqrt{(0)^2+(5)^2}+\sqrt{(6)^2+(-5)^2}+\sqrt{(-6)^2+(-5)^2}]*50 \text{ feet}\\d_{T}=[\sqrt{0+25}+\sqrt{36+25}+\sqrt{36+25}]*50 \text{ feet}\\d_{T}=[\sqrt{25}+\sqrt{61}+\sqrt{61}]*50 \text{ feet}\\d_{T}=[5+7.81+7.81]*50 \text{ feet}\\d_{T}=[20.62]*50 \text{ feet}\\d_{T}=1031 \text{ feet}\\

So the total distance travelled is 1031 feet.

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3 years ago
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