Problem 49 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In tests conducted by Auto Test Magazine on two identical models of the Phoenix Elite - one equipped with a standard engine and the other with a turbo-charger-it was found that the acceleration of the former is given by $$ a=f(t)=4+0.8 t \quad(0 \leq t \leq 12) $$ \(\mathrm{ft} / \mathrm{sec} / \mathrm{sec}, t\) sec after starting from rest at full throttle, whereas the acceleration of the latter is given by $$ a=g(t)=4+1.2 t+0.03 t^{2} \quad(0 \leq t \leq 12) $$ ft/sec/sec. How much faster is the turbo-charged model moving than the model with the standard engine at the end of a 10 -sec test run at full throttle?

Step-by-Step Solution

Answer

The turbo-charged model is moving 30 ft/sec faster than the model with the standard engine at the end of a 10-second test run at full throttle.

Step 1: Integrate the acceleration functions to find the velocity functions

Let's find the velocity functions for both models by integrating the acceleration functions: For the standard engine, the acceleration function is: \(a = f(t) = 4 + 0.8t\) Integrating this acceleration function with respect to t gives us the velocity function: \(v_{1}(t) = \int (4 + 0.8t) dt = 4t + 0.4t^2 + C_{1}\) For the turbo-charged engine, the acceleration function is: \(a = g(t) = 4 + 1.2t + 0.03t^2\) Integrating this acceleration function with respect to t gives us the velocity function: \(v_{2}(t) = \int (4 + 1.2t + 0.03t^2) dt = 4t + 0.6t^2 + 0.01t^3 + C_{2}\)

Step 2: Determine the constants of integration

Since both models start from rest (zero velocity) at t=0, we can plug t=0 into the velocity functions to solve for the constants of integration: For the standard engine, when t=0: \(0 = 4(0) + 0.4(0)^2 + C_{1} \Rightarrow C_{1} = 0\) For the turbo-charged engine, when t=0: \(0 = 4(0) + 0.6(0)^2 + 0.01(0)^3 + C_{2} \Rightarrow C_{2} = 0\) This gives us the velocity functions without the constants of integration: \(v_{1}(t) = 4t + 0.4t^2\) \(v_{2}(t) = 4t + 0.6t^2 + 0.01t^3\)

Step 3: Evaluate the velocity functions at t=10 seconds

Now we need to calculate the velocities at the end of the 10-second test run: For the standard engine, when t=10: \(v_{1}(10) = 4(10) + 0.4(10)^2 = 40 + 40 = 80\) For the turbo-charged engine, when t=10: \(v_{2}(10) = 4(10) + 0.6(10)^2 + 0.01(10)^3 = 40 + 60 + 10 = 110\)

Step 4: Calculate the difference in velocities

Finally, we'll calculate the difference in the velocities to determine how much faster the turbo-charged model is moving than the model with the standard engine at the end of the 10-second test run: Difference in velocities = \(v_{2}(10)\) - \(v_{1}(10)\) = 110 - 80 = 30 The turbo-charged model is moving 30 ft/sec faster than the model with the standard engine at the end of a 10-second test run at full throttle.

Key Concepts

Integration

Integration is a key concept in calculus, often used to find areas under curves or to determine accumulated quantities. In this exercise, integration helps us find the velocity functions from given acceleration functions. When we integrate an acceleration function with respect to time, we obtain a velocity function. This process involves finding the antiderivative of the acceleration function.

Think about it like this:

• The velocity function is essentially a collection of all the small changes (accelerations) happening over time.
• By integrating, we sum up all of these changes to see how fast an object is moving at any given time.

The constants of integration appear because an indefinite integral could start from any value, so we have to make sure the initial conditions are respected. In this problem, both starting velocities are zero, which simplifies our calculations.

Velocity function

The velocity function is an expression that tells us how fast an object is moving over time. In the context of the provided exercise, the velocity function is derived from the acceleration function by integration. Initially, when the object is at rest, its velocity is zero.

For example:

• For the standard engine model, the velocity function is derived as follows: \(v_{1}(t) = 4t + 0.4t^2\).
• For the turbo-charged engine model, the velocity function becomes: \(v_{2}(t) = 4t + 0.6t^2 + 0.01t^3\).

These velocity functions help us understand how the speed of each model changes over time, particularly by evaluating them at specific moments, such as at the 10-second mark.

Acceleration function

Acceleration functions describe how the velocity of an object changes over time. They provide the rate of change of the velocity and are crucial in understanding motion dynamics.

In our exercise:

• The acceleration function for the standard engine is: \(f(t) = 4 + 0.8t\).
• For the turbo-charged engine, it's given by: \(g(t) = 4 + 1.2t + 0.03t^2\).

Each function shows how different factors (like the presence of a turbo-charger) influence acceleration. These differences are pivotal when predicting how the velocities will accumulate over time when integrated.

Definite integrals

Definite integrals can be an incredibly useful tool when you need to calculate the exact accumulated change over a specific time interval. While the current problem focuses on indefinite integrals that don't specify such intervals, understanding definite integrals is still crucial.

• A definite integral gives the total accumulation of a quantity, like distance, over an interval \([a, b]\).
• It is often expressed as \(\int_{a}^{b} f(t) \, dt\), representing the area under the curve of a function between points \(a\) and \(b\).

This concept applies not only to physics problems but also to various real-world scenarios, such as calculating total consumption or growth over a period. In more advanced problems than ours, definite integrals would indicate precisely how much distance each car model covers over the test run.