The position of a particle moving along the x-axis at time t > 0 seconds is given by the function x(t) = e ^ t - 2t feet.

a) Find the average velocity of the particel over the interval [1,3].

b) In what direction and how fast is the particle moving at t= 1 seconds?

c) For what values of t is the particle moving to the right?

d) Find the postition of the particle when its velocity is 0.

x(t) = e ^ t - 2t

x(3) = e^3 - 6
x(1) = e - 2

avg speed = (e^3 - 6 - e + 2)/(3-1)
= (e^3 - e - 4)/2 ft/s

x'(t) = e^t - 2
x'(1) = e - 2 = about .72 ft/s
looks like it is moving to the right

moving to the right when e^t - 2 > 0
e^t > 2 , let's take ln of both sides
t ln e > ln2 ,
t > ln2

it stops when t = ln2
now find x(ln2) .

a) Average velocity over the interval [1,3] can be found by finding the displacement and dividing it by the time taken. So, let's calculate the displacement first:

x(3) - x(1) = (e^3 - 6) - (e - 2) feet

Now, let's calculate the time taken:

3 - 1 = 2 seconds

Average velocity = (e^3 - 6) - (e - 2) / 2 feet per second

b) At t = 1 second, we can find the velocity by taking the derivative of x(t):

x'(t) = e^t - 2

Substituting t = 1:

x'(1) = e^1 - 2 = e - 2 feet per second

So, the particle is moving in the e direction (exponential direction) at a speed of e - 2 feet per second.

c) The particle is moving to the right when its velocity is positive. So, let's find the values of t for which x'(t) > 0:

e^t - 2 > 0
e^t > 2
t > ln(2)

So, the particle is moving to the right for t > ln(2) seconds.

d) Velocity is the derivative of position, so when velocity is 0, we can set x'(t) = 0:

e^t - 2 = 0
e^t = 2
t = ln(2)

To find the position of the particle when its velocity is 0, we can substitute t = ln(2) back into the position function:

x(ln(2)) = e^(ln(2)) - 2ln(2) feet
x(ln(2)) = 2 - 2ln(2) feet

Therefore, the position of the particle when its velocity is 0 is 2 - 2ln(2) feet.

a) To find the average velocity of the particle over the interval [1,3], we need to calculate the change in position divided by the change in time. The formula for average velocity is:

average velocity = (x(3) - x(1)) / (3 - 1)

Plugging in the values into the given function:

x(3) = e^3 - 2*3 = e^3 - 6

x(1) = e^1 - 2*1 = e - 2

Therefore, the average velocity is:

average velocity = (e^3 - 6 - (e - 2)) / (3 - 1)
= (e^3 - e - 4) / 2

b) To find the direction and speed of the particle at t = 1 seconds, we need to find the velocity at that time. The velocity is the derivative of the position function x(t). Taking the derivative of x(t):

x'(t) = (d/dt) (e^t - 2t)
= e^t - 2

Plugging in t=1 into the velocity equation:

x'(1) = e^1 - 2
= e - 2

Therefore, at t = 1 seconds, the particle is moving in the direction of increasing x, and its speed is e-2 feet per second.

c) The particle is moving to the right when its velocity is positive. From the velocity equation, we can see that for values of t where e^t - 2 is greater than 0, the particle is moving to the right. Solving this inequality:

e^t - 2 > 0
e^t > 2
t > ln(2)

Therefore, the particle is moving to the right for values of t greater than ln(2).

d) To find the position of the particle when its velocity is 0, we need to set the velocity equation equal to 0 and solve for t:

e^t - 2 = 0
e^t = 2
t = ln(2)

Plugging this value of t back into the position function:

x(ln(2)) = e^(ln(2)) - 2*ln(2)
= 2 - 2ln(2)

Therefore, when the velocity is 0, the position of the particle is 2 - 2ln(2) feet.

a) To find the average velocity of the particle over the interval [1,3], we need to calculate the displacement (change in position) and divide it by the time interval.

Step 1: Calculate the displacement:
The position of the particle at time t=1 is x(1) = e^1 - 2(1) = e - 2 feet.
The position of the particle at time t=3 is x(3) = e^3 - 2(3) = e^3 - 6 feet.
The displacement between these two positions is Δx = x(3) - x(1) = (e^3 - 6) - (e - 2) = e^3 - e - 4 feet.

Step 2: Calculate the time interval:
The time interval is Δt = 3 - 1 = 2 seconds.

Step 3: Calculate the average velocity:
Average velocity = Δx / Δt = (e^3 - e - 4) / 2 feet per second.

Therefore, the average velocity of the particle over the interval [1,3] is (e^3 - e - 4) / 2 feet per second.

b) To determine the direction and speed of the particle at t=1 seconds, we need to analyze the sign of the velocity.

Step 1: Calculate the velocity:
The velocity of the particle at time t is given by the derivative of the position function x(t).

v(t) = dx(t) / dt = d/dt(e^t - 2t) = e^t - 2.

Step 2: Substitute t=1 into the velocity function:
v(1) = e^1 - 2 = e - 2.

Since the velocity v(1) = e - 2 is positive, the particle is moving in the positive direction (to the right) at t=1 seconds. The speed of the particle is e - 2 feet per second.

c) To determine when the particle is moving to the right, we need to analyze the sign of the velocity function.

Step 1: Recall the velocity function:
v(t) = e^t - 2.

Step 2: Set up the inequality to find when the velocity is positive (moving to the right):
e^t - 2 > 0.

Step 3: Solve for t:
e^t > 2.
t > ln(2),
where ln represents the natural logarithm.

Therefore, the particle is moving to the right for t > ln(2) seconds.

d) To find the position of the particle when its velocity is 0, we need to find the time(s) when the velocity function v(t) = e^t - 2 equals zero.

Step 1: Set up the equation:
e^t - 2 = 0.

Step 2: Solve for t:
e^t = 2.
t = ln(2),
where ln represents the natural logarithm.

Step 3: Substitute the value of t into the position function:
x(ln(2)) = e^(ln(2)) - 2(ln(2)) = 2 - 2ln(2) feet.

Therefore, the position of the particle when its velocity is 0 is 2 - 2ln(2) feet.