The position of a particle on the x-axis at time t, t > 0, is s(t) = ln(t) with t measured in seconds and s(t) measured in feet. What is the average velocity of the particle for e ≤ t ≤ 2e? A. ln2 B. the quotient of 1 and the quantity of 3 times C. the quotient of the natural logarithm of 2 and e D. the quotient of the natural logarithm of 2 and 2

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InesWalston InesWalston · Answer 1 · 2019-01-03T08:23:09+01:00

Answer:

C. The quotient of the natural logarithm of 2 and e.

Step-by-step explanation:

The position of a particle on the x-axis at time t, t > 0, is given by

[tex]s(t) = \ln(t)[/tex]

with t is in seconds and s(t) is in feet.

The rate of change is,

=[tex]\dfrac{f(b)-f(a)}{b-a}[/tex]

So rate of change of position or average velocity, for e ≤ t ≤ 2e will be,

[tex]=\dfrac{\ln2e-\ln e}{2e-e}[/tex]

[tex]=\dfrac{\ln2e-\ln e}{e}[/tex]

[tex]=\dfrac{\ln\frac{2e}{e}}{e}[/tex]

[tex]=\dfrac{\ln2}{e}[/tex]

Therefore, option C is the correct answer.

virtuematane virtuematane · Answer 2 · 2019-04-29T11:27:17+02:00

Answer:

Option: C is the correct answer.

C. the quotient of the natural logarithm of 2 and e .

Step-by-step explanation:

We are given a function s(t) that denotes the the position of a particle on the x-axis at time t, t > 0, as:

[tex]s(t)=\ln (t)[/tex]

Now we are asked to find the average velocity of the particle for e ≤ t ≤ 2e.

We know that the average velocity is defined as the ratio of total distance to total time.

Now total distance covered in e ≤ t ≤ 2e is:

[tex]s(2e)-s(e)[/tex]

[tex]=\ln (2e)-\ln (e)[/tex]

[tex]=ln (2e/e)[/tex]

( Since, [tex]\ln (m)-\ln (n)=\ln (m/n)[/tex] )

[tex]=\ln 2[/tex]

Also, total time is:

[tex]2e-e=e[/tex]

Hence, average velocity is:

[tex]Average\ Velocity=\dfrac{\ln (2)}{e}[/tex]

Option: C is the correct answer.