Question

[T] A car on the freeway accelerates according to \(a=15 \cos (\pi t)\), where \(t\) is measured in hours. Set up and solve the differential equation to determine the velocity of the car if it has an initial speed of 51 mph. After 40 minutes of driving, what is the driver's velocity?

Short answer

The driver's velocity after 40 minutes is approximately 54.12 mph.

Step-by-step solution

Understand the Problem

We are given an acceleration function \(a(t) = 15 \cos(\pi t)\) and we need to find the velocity \(v(t)\) given an initial speed of 51 mph. We are also asked to determine the velocity after 40 minutes.

Set Up the Differential Equation

Given \(a(t) = \frac{dv}{dt} = 15 \cos(\pi t)\), we need to integrate this to find the velocity \(v(t)\). We solve the differential equation \(\frac{dv}{dt} = 15 \cos(\pi t)\).

Integrate to Find the General Velocity Function

Integrate the acceleration function with respect to \(t\): \[ v(t) = \int 15 \cos(\pi t) \, dt \]. This gives us: \[ v(t) = \left(\frac{15}{\pi} \right) \sin(\pi t) + C \], where \(C\) is the constant of integration.

Apply the Initial Condition

The initial condition is that the velocity when \(t = 0\) is 51 mph. So substitute into the equation: \[ 51 = \left(\frac{15}{\pi}\right) \sin(0) + C \Rightarrow C = 51 \]. Therefore, the velocity function is \[ v(t) = \left(\frac{15}{\pi} \right) \sin(\pi t) + 51 \].

Convert Time to Hours

40 minutes is \(\frac{40}{60} = \frac{2}{3}\) hours. Substitute \(t = \frac{2}{3}\) into the velocity function to find \(v\left(\frac{2}{3}\right)\).

Calculate the Velocity After 40 Minutes

Using the velocity function \(v(t) = \left(\frac{15}{\pi} \right) \sin(\pi t) + 51 \), substitute \(t = \frac{2}{3}\): \[ v\left(\frac{2}{3}\right) = \left(\frac{15}{\pi}\right) \sin\left(\pi \times \frac{2}{3} \right) + 51 \]. Calculate \(\sin\left(\frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}\) to find \(v\left(\frac{2}{3}\right)\): \[ v\left(\frac{2}{3}\right) = \left(\frac{15}{\pi} \right) \times \frac{\sqrt{3}}{2} + 51 \approx 51 + \frac{15 \sqrt{3}}{2\pi} \].

Final Calculation

Calculate the final numerical value: \[ v\left(\frac{2}{3}\right) \approx 54.12 \] mph. Thus, after 40 minutes, the velocity is approximately 54.12 mph.

Key concepts

Velocity

Velocity is a measure of how fast something is moving in a particular direction. It is a vector quantity, which means it has both a magnitude (speed) and a direction. In the context of differential equations, we often denote velocity as the rate of change of position with respect to time, represented by the symbol \( v(t) \).
In our exercise, the challenge was to find the velocity of a car on the freeway given an initial speed. This involved solving a differential equation derived from the car's acceleration function.

• The initial speed (velocity at time \( t = 0 \)) was 51 mph.
• We aimed to find the velocity after 40 minutes of driving.

Knowing the initial speed allows us to determine the constant of integration when solving the differential equation.

Acceleration

Acceleration describes how quickly velocity is changing with time. It tells us whether an object is speeding up or slowing down, and the function given by the problem is \( a(t) = 15 \cos(\pi t) \). Since acceleration is the derivative of velocity, we denoted it here as \( \frac{dv}{dt} \).
Understanding acceleration is crucial to solving this type of problem, as it helps us set up the differential equation necessary for finding velocity.

• The form of this acceleration function indicates that it's periodic, as it uses the cosine function.
• This periodic nature impacts how the car's velocity changes over time, which must be considered when finding the velocity.

Integration

Integration is the process we use to find the original function from its derivative. In this exercise, we integrated the given acceleration function to find the velocity function.
The integral of \(15 \cos(\pi t)\) with respect to \( t \) gives us the general velocity function. Here's how it works:

• The integral \[ v(t) = \int 15 \cos(\pi t) \, dt \] results in\[ v(t) = \left(\frac{15}{\pi} \right) \sin(\pi t) + C \],where \(C\) is the constant of integration.
• This constant \(C\) is found by applying the initial condition.

Thus, integration is crucial in transitioning from acceleration to velocity.

Initial Conditions

Initial conditions are specific values that are provided to ensure the solutions to differential equations are tailored to the particular scenario.
They play a critical role in defining the constant of integration. In our scenario, the initial condition given was the velocity at \( t = 0 \) hours, which was 51 mph.

• Using this, the equation \[ 51 = \left(\frac{15}{\pi}\right) \sin(0) + C \] led us to determine that \( C = 51 \).
• Knowing the initial speed allowed us to construct a specific velocity function \[ v(t) = \left(\frac{15}{\pi} \right) \sin(\pi t) + 51 \].

Without initial conditions, we would have an infinite number of solutions to our differential equation.