For 4.95 seconds , a particle moves in a straight line according to the position function: s(t) = e^t(5-t)-5 .

a. when id the particle at rest ? when is particle moving forward ?
b. Find the total distance traveled by the particle in intervals [0,1] seconds and [1,2] seconds . which is larger and why ?
c. when was the particle speeding up ? slowing down ?

a. The particle is at rest when its velocity is equal to zero. To find when this occurs, we need to find the derivative of the position function:

s'(t) = e^t(5-t) - te^t - e^t

Setting this equal to zero and solving for t can give us when the particle is at rest.

As for when the particle is moving forward, we need to determine when its velocity is positive. This occurs when s'(t) > 0.

b. To find the total distance traveled by the particle in the intervals [0,1] seconds and [1,2] seconds, we need to calculate the definite integral of the absolute value of the velocity function over these intervals:

Distance traveled from 0 to 1 seconds:
∫[0,1] |s'(t)| dt

Distance traveled from 1 to 2 seconds:
∫[1,2] |s'(t)| dt

To determine which interval results in a larger distance, we can evaluate both integrals and compare the values.

c. The particle is speeding up when its acceleration is positive, and it is slowing down when its acceleration is negative. We can find the acceleration by taking the derivative of the velocity function, s'(t):

s''(t) = e^t(5 - 2t)

By examining the sign of s''(t), we can determine when the particle is speeding up or slowing down.

a. To find when the particle is at rest, we need to find when the velocity is zero. Velocity is the derivative of the position function. So let's find the velocity function.

1. Calculate the derivative of the position function:
s'(t) = [(e^t) * (5 - t)]' - 5' = e^t(5 - t) - e^t = e^t(5 - t - 1)

2. Set the velocity function equal to zero and solve for t:
e^t(5 - t - 1) = 0
e^t = 0 or (5 - t - 1) = 0

Since e^t cannot be zero, we focus on the second equation:
5 - t - 1 = 0
4 = t

So the particle is at rest at t = 4 seconds.

To determine when the particle is moving forward, we need to find when the velocity is positive.
e^t(5 - t - 1) > 0

We can break it down into two cases:
Case 1: e^t > 0 and (5 - t - 1) > 0
Case 2: e^t < 0 and (5 - t - 1) < 0

Case 1: e^t > 0 and 4 < t < 6
Case 2: This case is not possible since e^t cannot be negative.

Therefore, the particle is moving forward between t = 4 and t = 6 seconds.

b. To find the total distance traveled by the particle, we need to integrate the absolute value of the velocity function over the desired intervals.

1. For the interval [0,1] seconds:
The distance is given by the integral of the absolute value of the velocity function from 0 to 1:
∫[0,1] |e^t(5 - t - 1)| dt

2. For the interval [1,2] seconds:
The distance is given by the integral of the absolute value of the velocity function from 1 to 2:
∫[1,2] |e^t(5 - t - 1)| dt

To determine which interval has a larger total distance, we need to calculate the value of each integral. Unfortunately, it is not possible to evaluate these integrals analytically. You would need to use numerical methods, such as approximation techniques or software.

c. To find when the particle is speeding up or slowing down, we need to look at the acceleration. Since acceleration is the derivative of the velocity function, let's calculate it.

1. Calculate the derivative of the velocity function:
v'(t) = [(e^t(5 - t - 1))]' = (e^t(5 - t - 1))' = e^t(-t + 6)

2. Set the acceleration function equal to zero and solve for t:
e^t(-t + 6) = 0

Again, similar to before, e^t cannot be zero. So we focus on the second part of the equation:
-t + 6 = 0
t = 6

Therefore, the particle is speeding up at t = 6 seconds. To determine when it is slowing down, we can check the intervals around t = 6. However, since we do not have the exact values of the velocity function, we are unable to determine the exact points at which the particle is slowing down.

a. To determine when the particle is at rest and when it is moving forward, we need to find the velocity function by taking the derivative of the position function.

First, let's find the velocity function by differentiating the position function with respect to time:

v(t) = ds(t)/dt

To find ds(t)/dt, we can use the product rule of differentiation:

s(t) = e^t(5-t) - 5

ds(t)/dt = (d/dt)(e^t(5-t)) - (d/dt)(5)

Now, let's differentiate each term separately:

ds(t)/dt = [(d/dt)(e^t)](5-t) + e^t[(d/dt)(5-t)] - 0

The derivative of e^t is simply e^t, and the derivative of (5-t) is -1.

ds(t)/dt = e^t(5-t) - e^t

Now, to find when the particle is at rest, we need to find the values of t where the velocity, v(t), is zero:

v(t) = 0

e^t(5-t) - e^t = 0

Now, we can factor out e^t:

e^t[(5-t) - 1] = 0

Simplifying further:

e^t(4-t) = 0

To determine when the particle is at rest, we set e^t = 0 or (4-t) = 0:

e^t = 0 (No solutions, as e^t is always positive)

4 - t = 0
t = 4

Therefore, the particle is at rest when t = 4 seconds.

To determine when the particle is moving forward, we need to consider the sign of the velocity, v(t). If v(t) is positive for a given value of t, then the particle is moving forward.

From our earlier calculation of v(t):

v(t) = e^t(5-t) - e^t

We can observe that e^t is always positive, so the sign of v(t) is determined by the expression (5-t) - e^t.

For t < 5, (5-t) - e^t < 0, so v(t) < 0, meaning the particle is moving backward.

For t > 5, (5-t) - e^t > 0, so v(t) > 0, meaning the particle is moving forward.

Hence, the particle is moving forward when t > 5 seconds.

b. To find the total distance traveled by the particle in the intervals [0,1] seconds and [1,2] seconds, we need to determine the net distance traveled within those intervals.

The net distance is the absolute value of the displacement, which is the difference in position at the start and end of the interval.

For the interval [0,1] seconds:

The initial position is s(0) = e^0(5-0) - 5 = 0 - 5 = -5
The final position is s(1) = e^1(5-1) - 5 = e(5-1) - 5

To find the total distance traveled in this interval, we calculate the absolute value of the displacement:

Distance = |s(1) - s(0)|

For the interval [1,2] seconds:

The initial position is s(1) = e^1(5-1) - 5 = e(5-1) - 5
The final position is s(2) = e^2(5-2) - 5 = e^2(3) - 5

Again, we calculate the absolute value of the displacement:

Distance = |s(2) - s(1)|

To compare which interval has a larger distance traveled, we can evaluate the computed values of Distance for each interval.

c. To determine when the particle is speeding up or slowing down, we need to analyze the acceleration of the particle. The acceleration is the derivative of the velocity function, v(t).

Let's find the acceleration function by differentiating the velocity function:

a(t) = dv(t)/dt

dv(t)/dt = (d/dt)[e^t(5-t) - e^t]

Using the product rule of differentiation:

dv(t)/dt = [(d/dt)(e^t)](5-t) + e^t[(d/dt)(5-t)] - [(d/dt)(e^t)]

Differentiating each term:

dv(t)/dt = e^t(5-t) - e^t + e^t(-1) - 0

Simplifying:

dv(t)/dt = -t*e^t

The sign of the acceleration, a(t), will tell us if the particle is speeding up or slowing down.

For t > 0, -t is negative, and e^t is always positive. So, a(t), -t*e^t, is negative for t > 0.

This indicates that the particle is slowing down for t > 0.

Therefore, the particle is slowing down for all values of t greater than 0.