Question

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

Answer 1

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

$s(t) = \ln(t)$

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

The rate of change is,

=$\dfrac{f(b)-f(a)}{b-a}$

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

$=\dfrac{\ln2e-\ln e}{2e-e}$

$=\dfrac{\ln2e-\ln e}{e}$

$=\dfrac{\ln\frac{2e}{e}}{e}$

$=\dfrac{\ln2}{e}$

Therefore, option C is the correct answer.

Answer 2

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:

                              $s(t)=\ln (t)$

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:

$s(2e)-s(e)$

$=\ln (2e)-\ln (e)$

$=ln (2e/e)$

( Since, $\ln (m)-\ln (n)=\ln (m/n)$ )

$=\ln 2$

Also, total time is:

$2e-e=e$

Hence, average velocity is:

$Average\ Velocity=\dfrac{\ln (2)}{e}$

Option: C is the correct answer.