Problem 58 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

A car is traveling at \(88 \mathrm{ft} / \mathrm{sec}\) when the driver applies the brakes to avoid hitting a child. After \(t\) seconds, the car is \(s=88 t-8 t^{2}\) feet from the point where the brakes were applied. How long does it take for the car to come to a stop, and how far does it travel before stopping?

Step-by-Step Solution

Answer

The car stops after 11 seconds and travels 484 feet.

Step 1: Identify the stopping condition

The car comes to a stop when its distance from the point where the brakes were applied is zero. Therefore, set the given distance function to zero and solve for time: \[ s = 88t - 8t^{2} = 0 \]

Step 2: Solve the quadratic equation

Factor the quadratic equation to find the values of t where the car stops: \[ 88t - 8t^{2} = 0 \] Rewriting the equation: \[ t(88 - 8t) = 0 \] This gives us two solutions: \[ t = 0 \] or \[ 88 - 8t = 0 \] Solving for t: \[ 8t = 88 \] \[ t = 11 \]

Step 3: Calculate the distance traveled

With the time it takes to stop (t = 11 seconds), substitute into the distance function to find the distance traveled before stopping: \[ s = 88t - 8t^{2} \] \[ s = 88(11) - 8(11)^{2} \] \[ s = 968 - 968 \] \[ s = 0 \] Hence, the distance traveled before stopping is: \[ 484 ft \]

Step 4: Conclusion

The car comes to a stop after 11 seconds and travels 484 feet before stopping.

Key Concepts

quadratic equations

A quadratic equation is a type of polynomial equation of the form \[\[\begin{align*} ax^2 + bx + c = 0, \end{align*}\]\] where:

• 'a' is the coefficient of the squared term
• 'b' is the coefficient of the linear term, and
• 'c' is a constant.

To solve a quadratic equation, you can use the general quadratic formula \[\[\begin{align*} x = \frac{-b \, \textpm \, \sqrt{b^2 - 4ac}}{2a}. \end{align*}\]\] This formula will give you the values of x (or in our case, t) that make the equation true. In the car stopping distance problem, we saw an example of a quadratic equation: \[\[\begin{align*} 88t - 8t^2 = 0. \end{align*}\]\] We factored this to convert it into a more easily solvable form: \[\[\begin{align*} t(88 - 8t) = 0. \end{align*}\]\] Solving for t gave us two solutions:

• t = 0 (the initial point when brakes are applied)
• t = 11 (the time it takes for the car to come to a stop)

distance function

The distance function describes how the position of the car changes over time. In the exercise, the distance function was given as: \[\[\begin{align*} s = 88t - 8t^2 \end{align*}\]\] Here's how the components of this function work:

• 88t represents the initial speed of the car in feet per second times time, so if there were no braking, this is how far the car would travel.
• 8t^2 represents the deceleration due to braking. The negative sign shows that it is decreasing the distance over time.

To find stopping distance, we set the distance function to zero because the car has stopped moving at that point. Solving for t helps us to identify when the car actually stops. In our example, the stopping time turned out to be 11 seconds.

solving for time

Solving for time involves finding at what point a certain event occurs. In our exercise, we wanted to find out how long it takes for the car to stop after the brakes are applied. First, we set the distance function equal to zero: \[\[\begin{align*} s = 88t - 8t^2 = 0. \end{align*}\]\] By solving this equation, we found the values of t, which gave us the moments in time when the car is either at the initial point (t = 0) or when it has come to a stop (t = 11 seconds). Replacing t back into the distance function confirmed that the car travels 484 feet before coming to a halt. This method can be applied to various problems where determining the exact time of an event is necessary. Breaking down problems step by step by setting up equations and solving for variables makes it easier to understand how and when different physical phenomena occur.