Problem 121 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The constant deceleration required is approximately 16 ft/s².

Step 1: Solve the Differential Equation

We start with the differential equation \( \frac{d^2 s}{d t^2} = -k \) where \( k \) is a constant. To find \( \frac{d s}{d t} \), we integrate the equation once with respect to \( t \), yielding: \( \frac{d s}{d t} = -kt + C_1 \).Using the initial condition \( \frac{d s}{d t} = 88 \) when \( t = 0 \), we find \( C_1 = 88 \). Thus, \( \frac{d s}{d t} = -kt + 88 \).

Step 2: Integrate to Find s(t)

Next, we integrate \( \frac{d s}{d t} = -kt + 88 \) with respect to \( t \) to find \( s(t) \). This results in \( s(t) = -\frac{k}{2}t^2 + 88t + C_2 \).Applying the initial condition \( s = 0 \) when \( t = 0 \), we find \( C_2 = 0 \). Thus, \( s(t) = -\frac{k}{2}t^2 + 88t \).

Step 3: Find Time when the Velocity is Zero

To find the time \( t \) at which the car stops, set the velocity \( \frac{d s}{d t} = -kt + 88 = 0 \) and solve for \( t \). This gives: \[ t = \frac{88}{k} \].

Step 4: Solve for Constant Deceleration k

We need to find the value of \( k \) such that \( s = 242 \) when \( t = \frac{88}{k} \). Substitute \( t = \frac{88}{k} \) into \( s(t) = -\frac{k}{2}t^2 + 88t \) and set \( s = 242 \):\[ 242 = -\frac{k}{2}\left(\frac{88}{k}\right)^2 + 88\left(\frac{88}{k}\right) \].Simplify the equation:\[ 242 = \frac{88^2}{2k} \].Solving for \( k \), we find:\[ k = \frac{88^2}{2 \times 242} = \frac{7744}{484} \approx 16 \].

Key Concepts

Constant Deceleration

In physics, constant deceleration is a steady decrease in velocity over time. This contrasts with acceleration where velocity increases. When you slam on the brakes to stop a car, you essentially apply a force that decreases the car's speed steadily until it stops. In this scenario, we are trying to find the constant rate at which this deceleration happens.

Here’s an easy way to understand it:

• Constant deceleration means speed diminishes uniformly over time.
• It is expressed as a negative constant in equations since it's a reduction in velocity.
• Mathematically, it is represented in our differential equation as \(-k\), where \(k\) is the constant rate of deceleration.

When solving problems involving constant deceleration, knowing the initial velocity and distance to stop is crucial. That’s why, in this car problem, the initial speed (88 ft/sec) and stopping distance (242 ft) drive our calculations. By determining this constant, you understand how long it takes to halt a moving object safely.

Initial Value Problems

Initial value problems (IVPs) provide a useful way to find solutions to differential equations by specifying conditions at the start of the problem. Think of it like solving a puzzle: the initial values are the starting pieces that help guide you to the complete picture.

In the given problem, the IVP involves determining how the position changes over time using known initial conditions:

• The initial velocity of the car is \(88 \, \mathrm{ft/sec}\) when the brakes are applied.
• The car stops at \(s = 242 \, \mathrm{ft}\), which sets the final condition we need to meet.

This type of problem solves a differential equation step-by-step using these initial values:
- First, we determined how the velocity changes over time.
- Next, we integrated to find the position function.

The initial values guide each step, ensuring every part of the solution aligns with the conditions present right when you hit the brakes. They ensure the physics match the real-world scenario of braking a car.

Integration

Integration, a fundamental concept in calculus, involves finding a function given its rate of change. Here, it lets us transition from knowing how quickly something is slowing down to understanding exactly where it stops and when. Think of it as building a staircase: each step builds upon the previous.

In our mechanics problem, integration takes center stage in two main steps:

• First, integrating the initial differential equation allows us to express the velocity in terms of time: \(\frac{ds}{dt} = -kt + 88\).
• Second, a subsequent integration of this velocity function provides the formula for position over time: \(s(t) = -\frac{k}{2}t^2 + 88t\).

By integrating, you convert the initial problem about rates of change into actual values of stopping time and distance. This hands-on approach aids in connecting the theoretical mathematics to real-world events, like stopping a car safely on the road.