Problem 46 - Calculus an Applied Approach

Question

Stopping Distance A car is traveling at a rate of 66 feet per second (45 miles per hour) when the brakes are applied. The position function for the car is given by \(s=-8.25 t^{2}+66 t,\) where \(s\) is measured in feet and \(t\) is measured in seconds. Create a table showing the position, velocity, and acceleration for each given value of \(t .\) What can you conclude?

Step-by-Step Solution

Answer

The car's acceleration is a constant -16.5 ft/s² indicating a steady decrease in velocity (deceleration), until the car comes to stop. Exploration of specific values would require the creation of a table with selected values of 't' input into the 's', 'v', and 'a' functions.

Step 1: Derive the velocity function

The velocity of an object is the rate of change of its position with respect to time. It can be obtained by differentiating the position function with respect to time. Thus, the velocity function \(v(t)\) can be obtained as the derivative of \(s(t)\): \(v(t) = ds/dt = -16.5t + 66\).

Step 2: Derive the acceleration function

Acceleration is the rate of change of velocity with respect to time. It can be found by differentiating the velocity function. So, the acceleration function \(a(t)\) is the derivative of \(v(t)\): \(a(t) = dv/dt = -16.5\). In this case, acceleration is a constant.

Step 3: Create the table

Finally, you can plug in the values of time 't' into the position, velocity, and acceleration functions to get the corresponding values. This can be best represented in form of the table. Choose a range of values for 't' and calculate the values of 's', 'v', and 'a' for each.

Step 4: Conclusions

Given that the acceleration is constant and negative, this implies the car is decelerating. The velocity will continue to decrease until the car stops.

Key Concepts

Position Function

The position function of an object in motion describes its location at any given time. For this exercise, the position function is expressed as: \[ s(t) = -8.25t^2 + 66t \] where \( s \) is the position in feet and \( t \) is the time in seconds. This function is a quadratic equation, recognized by the presence of the \( t^2 \) term. The function indicates how the car's position changes over time as it slows down due to braking.Understanding the significance of each term and constant in a position function is crucial:

• The term \(-8.25t^2\) represents the acceleration due to braking, impacting the position in a negative quadratic fashion, meaning with time, the car slows down quadratically.
• The term \(66t\) indicates the initial impact of the car's velocity when brakes are applied, proportionally affecting the position over time.

By analyzing the position function, students can visualize how far the car travels before coming to a full stop, reinforcing their understanding of motion dynamics.

Velocity Function

The velocity function derives from the position function and represents the speed of the car over time. It's a crucial concept in calculus as it provides insight into how fast an object's position changes. To obtain the velocity function, we differentiate the position function \(s(t)\) with respect to time \(t\). The result is:\[ v(t) = \frac{ds}{dt} = -16.5t + 66 \]This linear equation reveals several key pieces of information:

• The coefficient \(-16.5\) in \(-16.5t\) represents the constant rate of deceleration, indicating that the car is slowing down.
• The constant \(66\) is the initial velocity in feet per second when the brakes are first applied.

The negative slope of the velocity function illustrates how the velocity decreases over time until it reaches zero, indicating the car has come to a stop.

Acceleration Function

Acceleration is a measure of how quickly velocity changes, and it's pivotal in understanding the motion dynamics better. It's found by differentiating the velocity function \(v(t)\) with respect to time \(t\). Here, since the velocity function is linear, its derivative is a constant:\[ a(t) = \frac{dv}{dt} = -16.5 \]This outcome reflects the consistent rate of deceleration. In simple terms:

• The acceleration function shows the rate at which the car is slowing down due to brakes.
• The negative sign indicates deceleration, meaning an opposing force is reducing the velocity.

Since the acceleration remains constant, it can be concluded that the brakes apply an unchanging force until the car fully stops. Understanding this concept is vital because it provides insights into the forces affecting moving objects, something essential not only in physics but also in real-world driving experiences.