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:
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:
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.