Question

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?

Short answer

The velocity and position of the car decrease as time increases (i.e., the car is slowing down). At the same time, the car's deceleration remains constant at -16.5 feet/sec\(^2\).

Step-by-step solution

Determining the Velocity of the Car

Find the first derivative of the position function, \(s(t) = -8.25t^2 + 66t\). The first derivative, \(v(t)\), will represent the velocity of the car. Using power rule for derivatives, \( \frac{d}{dx}[x^n] = nx^{(n-1)} \), the velocity function is \(v(t) = s'(t) = -16.5t + 66\).

Determining the Acceleration of the Car

Find the second derivative of the position function, which is the derivative of the velocity function. This will represent the acceleration of the car. Using power rule for derivatives again, the acceleration function is \(a(t) = v'(t) = -16.5\). This shows that the car is decelerating at a constant rate.

Creating the Table

Now, for different values of \(t\), we can find corresponding values of \(s(t)\), \(v(t)\), and \(a(t)\) using the functions we derived above and form a table. An example of this calculation for \(t=1\) second will be: \(s(1)=-8.25(1)^2 + 66(1)=57.75\) feet, \(v(1)=-16.5(1) + 66 = 49.5\) feet/sec and \(a(1)=-16.5\) feet/sec\(^2\). Similar calculations can be done for other values of \(t\).

Analyzing the Results

Analysing the results we can conclude that, as the time increases after the brakes are applied, the car slows down (velocity decreases) due to a constant deceleration and it covers lesser and lesser distance each second (position function reduces).

Key concepts

Position Function

When analyzing the movement of an object, such as a car, the position function gives us its location at any given time. In this case, the position function is expressed as \( s(t) = -8.25t^2 + 66t \). Here, \( s \) represents the position in feet and \( t \) stands for time in seconds.

The position function is derived from the concept of motion in physics. It shows how far an object has traveled from a starting point over time. This particular quadratic function suggests that the car's position does not increase linearly; instead, it starts slowing down over time.

When you apply an expression like this, you can determine the exact position of the car for any given time \( t \). For example, when \( t = 1 \) second, the position is 57.75 feet. Simply plug the value of \( t \) into the equation \( s(t) \) to find the position at that time. As time progresses, due to the negative \( -8.25t^2 \) term, the car continues to slow down.

Velocity

To understand how the car's speed changes over time, we look at the velocity function. This is found by taking the first derivative of the position function. In math, the derivative essentially measures how a function changes as its input changes.

For this car, the velocity function is \( v(t) = -16.5t + 66 \). Here's what it represents:

• \( v(t) \) is the velocity in feet per second at time \( t \).
• The car starts with an initial velocity of 66 feet per second when \( t = 0 \).
• For each passing second, the velocity decreases by 16.5 feet per second due to the \( -16.5t \) term.

In practical terms, this means the car is slowing down with each passing second after the brakes are applied. At \( t = 1 \) second, the velocity is 49.5 feet per second, and it keeps decreasing. This continuous decline in velocity reveals that the car doesn't move as fast as time goes on.

Acceleration

The concept of acceleration involves understanding how quickly the velocity of an object is changing. For the car in this scenario, we derive acceleration from the second derivative of the position function, or equivalently, the first derivative of the velocity function.

The acceleration function is \( a(t) = -16.5 \). This constant value indicates that the car experiences a consistent deceleration:

• Every second, the velocity decreases by 16.5 feet per second.
• Because acceleration is negative, it confirms that the car is slowing down steadily.

In this context, acceleration does not depend on time since it's a constant value of \( -16.5 \) feet/sec\(^2\). This steady deceleration reflects a linear and predictable decrease in the car's speed due to the applied brakes, indicating a uniform change in motion speed.