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sergejj [24]
3 years ago
9

An object moves with velocity v(t)=t^2-8t+7

Physics
1 answer:
solong [7]3 years ago
6 0
<span>a) write a polynomial expression for the position of the particle at any time t greater or equal to zero. 
</span>Position is found by integrating velocity: 

<span>s(t) = (t^3)/3 - 4t^2 + 7t + c 
</span>where c is a constant corresponding to the position at t=0. <span>

b) at what time(s) is the particle changing direction
</span>the particle changes direction whenever the velocity is zero; the velocity function equals 

<span>(t-1)(t-7) a difference of squares so the zeros are 1 and 7, it changes direction at 1 second and 7 seconds. </span><span>

c) find the total distance traveled by the particle from t=0 and t=4
</span><span>s(0) = c
s(1) = 8/3 + c
s(4) = 64/3 - 64 + 28 + c. 
</span>
from 0 to 1 the particle travels 8/3 units. From 1 to 4 it travels -(64/3 - 36 - 8/3) = (-(56/3 - 108/3)) 
<span>=-(-52/3) = 52/3 units 
</span>
<span>so in total it travels 52/3 + 8/3 =20 units</span>
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Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

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Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

\omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

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2 years ago
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There are no measurements shown in a table that accompanies
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