Problem 103 - Thomas Calculus

Question

\(\begin{array}{l}{\text { Stopping a car in time You are driving along a highway at a }} \\ {\text { steady } 60 \mathrm{mph}(88 \mathrm{ft} / \mathrm{sec}) \text { when you see an accident a head and }} \\ {\text { slam on the brakes. What constant deceleration is required to }} \\ {\text { stop your car in } 242 \mathrm{ft} \text { ? To find out, carry out the following }} \\ {\text { steps. }}\end{array}\) \(\begin{array}{l}{\text { 1. Solve the initial value problem }} \\ {\text { Differential equation: } \frac{d^{2} s}{d t^{2}}=-k \quad(k \text { constant })} \\ {\text { Initial conditions: } \frac{d s}{d t}=88 \text { and } s=0 \text { when } t=0}\end{array}\) \(\begin{array}{l}{\text { 2. Find the value of } t \text { that makes } d s / d t=0 . \text { (The answer will }} \\ {\text { involve } k . )} \\ {\text { 3. Find the value of } k \text { that makes } s=242 \text { for the value of } t} \\ {\text { you found in Step } 2 \text { . }}\end{array}\)

Step-by-Step Solution

Answer

The constant deceleration required is 16 ft/s².

Step 1: Solving the Differential Equation

The differential equation provided is \( \frac{d^{2}s}{dt^{2}} = -k \), which describes a constant deceleration. To solve it, integrate once with respect to \( t \) to get \( \frac{ds}{dt} = -kt + C_1 \). Given the initial condition \( \frac{ds}{dt} = 88 \) when \( t = 0 \), substituting these values gives \( 88 = C_1 \). Therefore, \( \frac{ds}{dt} = -kt + 88 \).

Step 2: Integrating to Find the Position Function

Integrate the velocity function \( \frac{ds}{dt} = -kt + 88 \) again with respect to \( t \) to find the position function \( s(t) = -\frac{k}{2}t^{2} + 88t + C_2 \). Using the initial condition \( s = 0 \) when \( t = 0 \) gives \( 0 = 0 + 0 + C_2 \). Therefore, \( C_2 = 0 \) and the position function becomes \( s(t) = -\frac{k}{2}t^{2} + 88t \).

Step 3: Finding the Time When the Car Stops

To find the time \( t \) when the car stops, set \( \frac{ds}{dt} = 0 \): \(-kt + 88 = 0\). Solving for \( t \) gives \( t = \frac{88}{k} \). This is the time it takes for the car to stop moving.

Step 4: Calculating the Deceleration k

To find \( k \), use the condition \( s = 242 \) when \( t = \frac{88}{k} \). Substitute these into the position function: \( 242 = -\frac{k}{2}\left(\frac{88}{k}\right)^{2} + 88\left(\frac{88}{k}\right) \). Simplifying this gives: \[ 242 = -\frac{7744}{2k} + \frac{7744}{k} \], which simplifies further to \( \frac{7744}{2k} = 242 \). Solving for \( k \), we get \( k = \frac{7744}{484} \) which equals \( 16 \).

Step 5: Confirming the Solution

Re-check the calculations by substituting \( k = 16 \) and \( t = 5.5 \) in the position function to see if \( s = 242 \). Thus, \( s(5.5) = -\frac{16}{2} \times (5.5)^2 + 88 \times 5.5 = 242 \), confirming \( k = 16 \) is correct.

Key Concepts

Initial Value Problem

An Initial Value Problem (IVP) in the context of differential equations involves finding a function that not only solves the differential equation but also satisfies specific conditions at a designated point. Here, the problem is to determine the behavior of the car as it comes to a stop, starting from given conditions.

In this exercise, we are given a second-order differential equation \( \frac{d^2s}{dt^2} = -k \), which models the deceleration as a constant value. The initial conditions provided are the car's initial velocity \( \frac{ds}{dt} = 88 \) ft/s (since the car moves at 60 mph) and the position \( s = 0 \) at \( t = 0 \).

Solving an initial value problem typically involves several steps:

• First, integrating the differential equation to find the velocity function.
• Second, using the initial conditions to solve for any constants of integration.
• Finally, integrating again if required to find the position function or other aspects of the object’s motion.

These steps help form a complete picture of the object’s motion from the start.

Velocity Function

The velocity function is crucial as it tells us how fast the object is moving at any point in time. In this exercise, once the initial differential equation is integrated, we obtain the velocity function: \( \frac{ds}{dt} = -kt + 88 \).

This function shows that the car's speed reduces linearly over time due to the constant deceleration. The term \(-kt\) signifies a reduction in speed over time, while the constant 88 indicates the initial velocity of the car.

Understanding the Role of the Velocity Function

The velocity function helps us to determine when the car comes to a stop. Since we seek the time when the velocity is zero, we set \( \frac{ds}{dt} = 0 \), solving for \( t \) to find \( t = \frac{88}{k} \).

• The velocity function is key to understanding how long it takes for the object to stop.
• It also aids in finding the required deceleration to halt the object precisely at a given distance.

This approach is commonplace when examining problems with constant acceleration or deceleration.

Deceleration

Deceleration is simply reducing speed and is considered a negative acceleration. In problems like this, it is essential because it determines how rapidly the car can come to rest. The differential equation \( \frac{d^2s}{dt^2} = -k \) indicates a constant deceleration with magnitude \( k \).

Calculating Deceleration

To solve for the deceleration \( k \), we substitute the conditions back into our derived functions. Using the known fact that the car must stop after 242 feet, we solve for \( k \) using the equation \[ 242 = -\frac{k}{2} \left(\frac{88}{k}\right)^2 + 88 \left(\frac{88}{k}\right) \].

Simplifying yields \( k = 16 \). This calculation shows how a linear adjustment to the rate (velocity function) translates to a feasible stop point with constant deceleration.

Position Function

The position function describes how an object's position changes over time, providing insight into its entire journey. Starting from the velocity function \( \frac{ds}{dt} = -kt + 88 \), we integrate this to obtain the position function.

The integration results in \( s(t) = -\frac{k}{2}t^{2} + 88t \). This quadratic function reflects the path traveled by the car, showing its displacement over time.

The position function is especially crucial for confirming that the desired stopping distance is achieved. By substituting \( t = \frac{88}{k} \) and examining if \( s(242) \) is satisfied, we ensure that our function accurately predicts real-world scenarios, like stopping precisely 242 feet ahead.

• The position function provides visualization of the object's travel, making comprehension easier.
• It acts as a check to validate the velocity function and deceleration used in the calculations.