Problem 155 - Calculus Volume 1

Question

The position function \(s(t)=t^{2}-3 t-4\) represents the position of the back of a car backing out of a driveway and then driving in a straight line, where s is in feet and t is in seconds. In this case, s(t) = 0 represents the time at which the back of the car is at the garage door, so s(0) = ?4 is the starting position of the car, 4 feet inside the garage. a. Determine the velocity of the car when s(t) = 0. b. Determine the velocity of the car when s(t) = 14.

Step-by-Step Solution

Answer

a. Velocity is 5 ft/s for s(t)=0. b. Velocity is 9 ft/s for s(t)=14.

Step 1: Find the Velocity Function

The velocity of the car is the derivative of the position function with respect to time. The position function is given by: \[ s(t) = t^2 - 3t - 4 \]To find the velocity function, we differentiate \( s(t) \):\[ v(t) = \frac{d}{dt}(t^2 - 3t - 4) \]\[ v(t) = 2t - 3 \] Thus, the velocity function is \( v(t) = 2t - 3 \).

Step 2: Calculate Time When s(t) = 0

To find the time \( t \) when \( s(t) = 0 \), we solve the equation:\[ t^2 - 3t - 4 = 0 \]Apply the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -3, c = -4 \):\[ t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]\[ t = \frac{3 \pm \sqrt{9 + 16}}{2} \]\[ t = \frac{3 \pm 5}{2} \]So, \( t_1 = 4 \) and \( t_2 = -1 \) (we discard \( t_2 = -1 \) since time cannot be negative). Thus \( t = 4 \) when \( s(t) = 0 \).

Step 3: Calculate Velocity when s(t) = 0

Now, substitute \( t = 4 \) into the velocity function:\[ v(4) = 2(4) - 3 \]\[ v(4) = 8 - 3 \]\[ v(4) = 5 \]The velocity of the car when \( s(t) = 0 \) is 5 feet per second.

Step 4: Calculate Time When s(t) = 14

Solve for \( t \) when \( s(t) = 14 \) using the equation:\[ t^2 - 3t - 4 = 14 \]Simplify to:\[ t^2 - 3t - 18 = 0 \]Use the quadratic formula with \( a = 1, b = -3, c = -18 \):\[ t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-18)}}{2 \cdot 1} \]\[ t = \frac{3 \pm \sqrt{9 + 72}}{2} \]\[ t = \frac{3 \pm \sqrt{81}}{2} \]\[ t = \frac{3 \pm 9}{2} \]\( t_1 = 6 \) and \( t_2 = -3 \) (discard \( t_2 = -3 \) because time must be positive). Thus \( t = 6 \) when \( s(t) = 14 \).

Step 5: Calculate Velocity when s(t) = 14

Substitute \( t = 6 \) into the velocity function:\[ v(6) = 2(6) - 3 \]\[ v(6) = 12 - 3 \]\[ v(6) = 9 \]The velocity of the car when \( s(t) = 14 \) is 9 feet per second.

Key Concepts

Velocity Function

The velocity function describes how quickly the position of an object is changing over time. In our exercise, the position function given is \( s(t) = t^2 - 3t - 4 \), representing the car's position at any time \( t \). To find the velocity function, we calculate the derivative of the position function with respect to time. This process essentially measures the steepness or rate of change of the car's position.

The derivative of the position function, \( s(t) \), gives us the velocity function, \( v(t) = 2t - 3 \). This function tells us the instantaneous velocity of the car at any time \( t \). Meaning, if we plug a value of \( t \) into the velocity function, we get how fast the car is moving at that particular moment in time.

Derivative

Understanding derivatives is key to grasping the concept of velocity. A derivative indicates how one quantity changes as another quantity changes. In simpler terms, if you have a position function, taking its derivative will reveal how fast that position is changing — its velocity.

To compute the derivative of \( s(t) = t^2 - 3t - 4 \), we use basic differentiation rules:

• The derivative of \( t^2 \) is \( 2t \), applying the power rule.
• The derivative of \(-3t\) is \(-3\), since the coefficient remains constant.
• The derivative of a constant \(-4\) is zero because constants do not change.

Combining these, we obtain the velocity function \( v(t) = 2t - 3 \). Thus, derivatives help in transforming position information into velocity insight.

Quadratic Formula

The quadratic formula is a valuable tool for finding the values of \( t \) when a quadratic equation equals zero. Our position function \( s(t) \) leads us to solve quadratic equations to find when the car reaches certain positions.

The general form is \( ax^2 + bx + c = 0 \), and the quadratic formula is given as:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our step-by-step solution, solving \( t^2 - 3t - 4 = 0 \) for \( s(t) = 0 \) gives:

• \( a = 1 \), \( b = -3 \), \( c = -4 \)
• Substituting into the formula provides two values of \( t \): \( t_1 = 4 \) and \( t_2 = -1 \) (discarded).

The quadratic formula helped us identify that the car's back is at the garage door when \( t = 4 \) seconds.

Solving Equations

Solving equations is a central part of understanding the car's motion. We determine when certain events occur by setting the position function \( s(t) \) equal to specific values.

In the exercise:

• To find when \( s(t) = 0 \) (car at the garage), we solve \( t^2 - 3t - 4 = 0 \).
• To find when \( s(t) = 14 \) (car 14 feet away from the garage), we solve \( t^2 - 3t - 18 = 0 \).

Both require the use of the quadratic formula for solving, yielding specific times \( t = 4 \) and \( t = 6 \), respectively. The solutions reveal when the car reaches these positions, enabling us to substitute back into the velocity function to find its speed at these times.