The velocity function (in meters per second) is given for a particle moving along a line.

v(t) = 3t − 7, 0 ≤ t ≤ 3
(a) Find the displacement.
-7.5 m
(b) Find the distance traveled by the particle during the given time interval.

negative v at start, positive at 3 so reverses direction during the 3 seconds at t = 7/3

v = 3 t - 7
for vector displacement
x = (3/2)t^2 - 7 t + c
at t = 3, x = 27/2 - 21 + c
at t = 0, x = c
so
displacement = -15/2 = -7.5 agree
then
for distance
integrate from t = 0 to t = 7/3
x = (3/2)t^2 - 7 t + c = (3/2)(49/9) - 49/3 = -49 /6 , negative motion
then integrate from t = 7/3 to t = 3
at t= 3
x = 27/2 - 21 + c
at t = 7/3
x = (3/2)(49/9) -49/3+ c = -49/6 +c
motion = 27/2 -21+49/6 = + 2/3 positive motion
total distance moved = 49/6 + 4/6 = 53/6

Why did the particle travel to the disco? Because it wanted to groove to the velocity function! Let's calculate its displacement and distance traveled.

(a) To find the displacement, we need to integrate the velocity function over the given time interval. So, let's find the antiderivative of v(t):
∫(3t - 7) dt = (3/2)t² - 7t + C

Next, we evaluate the antiderivative at the upper and lower limits of the interval:
Displacement = [(3/2)(3)² - 7(3)] - [(3/2)(0)² - 7(0)]
Displacement = [(3/2)(9) - 21] - [(3/2)(0) - 0]
Displacement = [27/2 - 21] - [0 - 0]
Displacement = -21/2
Displacement = -10.5 meters

So, the particle's displacement is -10.5 meters.

(b) To find the distance traveled, we need to consider the absolute value of the velocity function, as distance cannot be negative.

Let's split the given time interval into two parts:
0 ≤ t ≤ 2, where v(t) = 3t - 7
2 ≤ t ≤ 3, where v(t) = 3t - 7

For the first part:
Distance1 = ∫|3t - 7| dt = ∫(7 - 3t) dt = [7t - (3/2)t²]ₒᵥ₂ = [(7)(2) - (3/2)(2)²] - [(7)(0) - (3/2)(0)²]
Distance1 = [14 - 6] - [0 - 0] = 8 meters

For the second part:
Distance2 = ∫|3t - 7| dt = ∫(3t - 7) dt = [(3/2)t² - 7t]₂₃ = [(3/2)(3)² - 7(3)] - [(3/2)(2)² - 7(2)]
Distance2 = [(3/2)(9) - 21] - [(3/2)(4) - 14] = 27/2 - 21 - 6 + 14 = 5.5 meters

Therefore, the total distance traveled by the particle during the given time interval is:
Distance = Distance1 + Distance2 = 8 + 5.5 = 13.5 meters.

So, the particle traveled a total distance of 13.5 meters. Hope it didn't get too tired from all that dancing!

To find the displacement of the particle, we need to integrate the velocity function over the given time interval.

The displacement can be found using the formula:

Displacement = ∫[a to b] v(t) dt

Where v(t) is the velocity function and [a, b] is the time interval.

In this case, the velocity function is given as v(t) = 3t - 7, and the time interval is 0 ≤ t ≤ 3.

We can integrate v(t) to find the displacement:

∫[0 to 3] (3t - 7) dt

Integrating, we get:

Displacement = ½ (3t^2 - 7t) │[0 to 3]
= ½ [(3(3)^2 - 7(3)) - (3(0)^2 - 7(0))]
= ½ [(27 - 21) - 0]
= ½ (6)
= 3 meters

Therefore, the displacement of the particle is 3 meters.

Now, to find the distance traveled by the particle during the given time interval, we need to consider the absolute value of the displacement. Since distance is always positive:

Distance = |Displacement| = |3| = 3 meters

Therefore, the distance traveled by the particle during the given time interval is 3 meters.

To find the displacement, we need to calculate the change in position of the particle. We can do this by integrating the velocity function over the given time interval.

(a) Displacement:
To find the displacement, we need to integrate the velocity function from the initial time (0) to the final time (3).

∫(0 to 3) (3t - 7) dt

First, we calculate the integral of 3t:

∫(0 to 3) 3t dt = (3/2) t^2 | (0 to 3)

Plugging in the values, we get:

[(3/2)(3)^2] - [(3/2)(0)^2]
= (3/2)(9) - (3/2)(0)
= 27/2

Next, we calculate the integral of -7:

∫(0 to 3) -7 dt = -7t | (0 to 3)

Plugging in the values, we get:

[-7(3)] - [-7(0)]
= -21

Now, we can calculate the displacement by subtracting the second integral from the first:

Displacement = (27/2) - (-21)
= 27/2 + 21/1
= (27 + 42)/2
= 69/2
= 34.5 m

Therefore, the displacement is 34.5 meters.

(b) Distance traveled:
To find the distance traveled by the particle, we need to consider the absolute value of the velocity function.

The absolute value of the velocity function, |v(t)| = |3t - 7|, gives us the speed at any given point.

To find the distance traveled, we need to calculate the integral of |v(t)| over the given time interval.

∫(0 to 3) |3t - 7| dt

Since the velocity function is a linear function, we can split the integral into two parts, depending on the sign of 3t - 7.

When 3t - 7 is positive (3t > 7):
∫(0 to 3) (3t - 7) dt

We already found this integral earlier, and the result was 27/2.

When 3t - 7 is negative (3t < 7):
∫(0 to 3) (7 - 3t) dt

∫(0 to 3) 7 - 3t dt = 7t - (3/2)t^2 | (0 to 3)

Plugging in the values, we get:

[7(3) - (3/2)(3)^2] - [7(0) - (3/2)(0)^2]
= 21 - (9/2)
= 33/2

Now, we can calculate the total distance traveled by adding the two results together:

Distance traveled = (27/2) + (33/2)
= (27 + 33)/2
= 60/2
= 30 m

Therefore, the distance traveled by the particle during the given time interval is 30 meters.