Problem 99 - Thomas Calculus in SI Units

Question

Stopping a car in time You are driving along a highway at a steady \(108 \mathrm{km} / \mathrm{h}(30 \mathrm{m} / \mathrm{s})\) when you see an accident ahead and slam on the brakes. What constant deceleration is required to stop your car in \(75 \mathrm{m} ?\) 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: \(\quad \frac{d s}{d t}=30\) and \(s=0\) when \(t=0\) 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 \(s=75\) for the value of \(t\) you found in Step 2.

Step-by-Step Solution

Answer

The deceleration required is \( 6 \text{ m/s}^2 \).

Step 1: Solve the Differential Equation

We start with the differential equation \( \frac{d^2 s}{dt^2} = -k \) where \( k \) is a constant. Integrating both sides with respect to \( t \), we get \( \frac{ds}{dt} = -kt + C_1 \). From the initial condition \( \frac{ds}{dt} = 30 \) when \( t = 0 \), we find \( C_1 = 30 \). Thus, \( \frac{ds}{dt} = -kt + 30 \).

Step 2: Further Integration to Find s(t)

Integrating \( \frac{ds}{dt} = -kt + 30 \) with respect to \( t \) gives \( s(t) = -\frac{k}{2}t^2 + 30t + C_2 \). Using the initial condition \( s = 0 \) when \( t = 0 \), we find \( C_2 = 0 \). Thus, \( s(t) = -\frac{k}{2}t^2 + 30t \).

Step 3: Find the Time When the Car Stops

We need \( \frac{ds}{dt} = 0 \) to find when the car stops. From \( \frac{ds}{dt} = -kt + 30 \), setting it to zero gives \( -kt + 30 = 0 \). Solving for \( t \), we get \( t = \frac{30}{k} \).

Step 4: Determine the Value of k

We find \( k \) by ensuring \( s = 75 \) meters when \( t = \frac{30}{k} \). Substitute \( t = \frac{30}{k} \) into \( s(t) = -\frac{k}{2}t^2 + 30t \) to get \( 75 = -\frac{k}{2} \left( \frac{30}{k} \right)^2 + 30 \left( \frac{30}{k} \right) \). Simplifying, \( 75 = \frac{900}{2k} \). Rearranging, \( 150k = 900 \), thus \( k = 6 \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. These are powerful tools in modeling situations where the rate of change of a quantity is known. In this exercise, we looked at a second-order differential equation, which takes the form \( \frac{d^2 s}{dt^2} = -k \), representing uniform deceleration. This means the rate at which the car's speed decreases is constant. The negative sign indicates deceleration (slowing down).
Using differential equations, we usually solve for unknown functions, which in this case are velocity and position. Integrating the differential equation progressively gives us velocity and then position concerning time. This way, we explore how the car's speed and position change over time after applying brakes.

Initial Value Problems

In mathematics, an initial value problem involves determining a function that satisfies a differential equation and an initial set of conditions. Initial value problems are common in physics when modeling real-world phenomena that involve changing rates. For this exercise, specific initial conditions were given: the car's velocity is 30 m/s at the moment brakes are applied (time \( t = 0 \)) and the initial position (\( s = 0 \)).
These conditions are essential to finding a unique solution to the differential equations. By plugging these initial values into our equations, we solve for constants that appear during integration, thus tailoring the general solution of the differential equation to our specific problem.

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the causes of motion. It's focused on variables like displacement, velocity, and acceleration. In this exercise, kinematics comes into play as we analyze the car's motion when stopping.
The vehicle's motion equations are derived from the given constant deceleration. With known initial speed and required stopping distance, we determine the car's path. Using equations of motion derived from initial values and conditions, such as \( s(t) = -\frac{k}{2}t^2 + 30t \), we provide insights into the car's behavior. This helps in predicting when the car will stop and under what conditions.

Physics Problem Solving

Physics problem-solving often involves breaking down complex motions and interactions into simpler, solvable parts. In this vehicle deceleration problem, the work begins by understanding the physics behind uniform deceleration and applying mathematical tools.
Steps involved include formulating a differential equation based on known relationships among motion variables, solving it in stages via integration, using provided initial conditions to determine constants, and finally substituting back to conform to the target conditions — like fully stopping within 75 meters. By following these logical and stepwise processes, the problem is decomposed into manageable parts, leading to the determination of the deceleration \( k = 6 \), which fulfills the initial problem requirements.