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 1 ≤ t ≤ e?
a)1
b)e
c)e-1
d)1/(e-1)
s(1) = 0
s(e) = 1
so distance traveled in (e - 1) seconds is one foot
v = 1/(e-1) = .582 ft/s
d)
come on foo
answer the question
Well, let's find the average velocity using clown math!
First, we need to find the displacement of the particle. To do that, we'll find s(e) - s(1):
s(e) = ln(e) = 1
s(1) = ln(1) = 0
So, the displacement is 1 - 0 = 1 foot.
Next, we'll find the time interval:
t(e) = e seconds
t(1) = 1 second
So, the time interval is e - 1 seconds.
Finally, average velocity is displacement over time, so:
Average velocity = 1 foot / (e - 1) seconds
Therefore, the answer is d) 1/(e-1).
But remember, clown math should never be taken too seriously! 😄🤡
To find the average velocity of the particle for 1 ≤ t ≤ e, we need to calculate the change in position divided by the change in time.
The change in position, Δs, is given by s(e) - s(1), and the change in time, Δt, is e - 1.
Δs = s(e) - s(1)
= ln(e) - ln(1)
= 1 - 0
= 1
Δt = e - 1
Therefore, the average velocity, v, is given by v = Δs/Δt = 1/(e - 1).
So, the correct answer is d) 1/(e - 1).
To find the average velocity of the particle for 1 ≤ t ≤ e, we need to calculate the change in position (Δs) divided by the change in time (Δt) within this interval.
Let's start by finding the change in position (Δs) and change in time (Δt). Substituting the given limits, we have:
Δs = s(e) - s(1)
Δt = e - 1
Now, let's calculate the change in position (Δs):
Δs = ln(e) - ln(1)
= 1 - 0
= 1
Next, let's calculate the change in time (Δt):
Δt = e - 1
Now, we can find the average velocity (Vavg) by dividing the change in position by the change in time:
Vavg = Δs/Δt
= 1/(e - 1)
Therefore, the average velocity of the particle for 1 ≤ t ≤ e is 1/(e - 1). Thus, the correct answer is (d) 1/(e - 1).