The given question asks about a car's motion based on the position equation $s = 5t - 5t^2$. Let's break this down and solve each part.

Given:

The position of the car is described by the equation: $s = 5t - 5t^2$ where $s$ is in meters and $t$ is in seconds.

Questions:

1. (a) Find the displacement of the car between $t = 3 \, \text{s}$ and $t = 5 \, \text{s}$.
2. (b) Find the average velocity during the same interval.
3. (c) Find the instantaneous velocity at the given times $t = 3 \, \text{s}$ and $t = 5 \, \text{s}$.

(a) Displacement:

The displacement is the change in position between $t_1 = 3$ and $t_2 = 5$.

Substitute $t_1 = 3$ into the position equation: $s_1 = 5(3) - 5(3)^2 = 15 - 45 = -30 \, \text{m}.$

Substitute $t_2 = 5$ into the position equation: $s_2 = 5(5) - 5(5)^2 = 25 - 125 = -100 \, \text{m}.$

The displacement is: $\Delta s = s_2 - s_1 = -100 - (-30) = -100 + 30 = -70 \, \text{m}.$

(b) Average Velocity:

The average velocity is defined as: $v_{\text{avg}} = \frac{\Delta s}{\Delta t}.$

From part (a), $\Delta s = -70 \, \text{m}$, and $\Delta t = t_2 - t_1 = 5 - 3 = 2 \, \text{s}$.

$v_{\text{avg}} = \frac{-70}{2} = -35 \, \text{m/s}.$

(c) Instantaneous Velocity:

The instantaneous velocity is the derivative of the position function: $v(t) = \frac{ds}{dt} = \frac{d}{dt}(5t - 5t^2).$

Differentiate $s = 5t - 5t^2$: $v(t) = 5 - 10t.$

At $t = 3$: $v(3) = 5 - 10(3) = 5 - 30 = -25 \, \text{m/s}.$

At $t = 5$: $v(5) = 5 - 10(5) = 5 - 50 = -45 \, \text{m/s}.$

Final Answers:

1. Displacement: $-70 \, \text{m}$.
2. Average Velocity: $-35 \, \text{m/s}$.
3. Instantaneous Velocities:
   • At $t = 3$: $-25 \, \text{m/s}$.
   • At $t = 5$: $-45 \, \text{m/s}$.

Would you like additional clarification on any part? Here are 5 related questions for further practice:

1. What is the total distance traveled by the car between $t = 3 \, \text{s}$ and $t = 5 \, \text{s}$?
2. Find the acceleration of the car as a function of time.
3. Determine the time at which the car comes to rest (i.e., $v(t) = 0$).
4. How would the displacement change if the motion occurred over the interval $t = 0 \, \text{s}$ to $t = 3 \, \text{s}$?
5. Plot the position, velocity, and acceleration functions over the interval $t = 0 \, \text{s}$ to $t = 5 \, \text{s}$.

Tip: Remember, displacement considers the net change in position, while distance accounts for the total path traveled!