Problem 34 - Brief Applied Calculus

Question

Velocity After \(t\) hours a passenger train is \(s(t)=24 t^{2}-2 t^{3}\) miles due west of its starting point (for \(0 \leq t \leq 12\) ). a. Find its velocity at time \(t=4\) hours. b. Find its velocity at time \(t=10\) hours. c. Find its acceleration at time \(t=1\) hour.

Step-by-Step Solution

Answer

a. 96 mph, b. -120 mph, c. 36 mph²

Step 1: Understand the Problem

We are given that a train's position, after time \(t\) hours, is defined by the function \(s(t)=24t^2-2t^3\). We need to find the velocity at specific times \(t=4\) and \(t=10\), and the acceleration at \(t=1\).

Step 2: Find the Velocity Function

Velocity is the first derivative of the position function with respect to time. Given \(s(t) = 24t^2 - 2t^3\), we find its derivative, \(v(t) = \frac{ds}{dt} = \frac{d}{dt}(24t^2 - 2t^3) = 48t - 6t^2\).

Step 3: Calculate Velocity at t=4

Substitute \(t = 4\) into the velocity function to find the velocity at this time.\[v(4) = 48(4) - 6(4)^2 = 192 - 96 = 96\]Thus, the velocity at \(t = 4\) hours is 96 miles per hour.

Step 4: Calculate Velocity at t=10

Substitute \(t = 10\) into the velocity function to find the velocity at this time.\[v(10) = 48(10) - 6(10)^2 = 480 - 600 = -120\]Thus, the velocity at \(t = 10\) hours is -120 miles per hour.

Step 5: Find the Acceleration Function

Acceleration is the derivative of the velocity function. From the velocity function \(v(t) = 48t - 6t^2\), we find its derivative, \(a(t) = \frac{dv}{dt} = \frac{d}{dt}(48t - 6t^2) = 48 - 12t\).

Step 6: Calculate Acceleration at t=1

Substitute \(t = 1\) into the acceleration function to find the acceleration at this time.\[a(1) = 48 - 12(1) = 48 - 12 = 36\]Thus, the acceleration at \(t = 1\) hour is 36 miles per hour squared.

Key Concepts

Velocity

In the context of differential calculus, velocity gives us insight into an object's speed and direction at a specific time. For a given function representing the position of an object over time, like in this example with the train, velocity is found by deriving the position function with respect to time. Simply put, the velocity shows us how fast the position changes.

To find the velocity, we take the first derivative of the position function. Here, our position function is described as \( s(t) = 24t^2 - 2t^3 \). By deriving, we get the velocity function \( v(t) = 48t - 6t^2 \).

This formula means at any time \( t \), the train’s velocity could be calculated. For instance:

• At \( t = 4 \) hours, the train moves at \( 96 \) miles per hour.
• At \( t = 10 \) hours, we notice the train is moving in the opposite direction at \( -120 \) miles per hour. The negative sign indicates a reversal in direction.

This demonstrates how velocity not only measures speed but also ties in direction, allowing for a comprehensive understanding of motion.

Acceleration

Acceleration adds another layer to understanding motion, describing how velocity changes over time. In our exercise, once we have the velocity function \( v(t) = 48t - 6t^2 \), finding acceleration is a matter of deriving velocity with respect to time.

The acceleration function turns out to be \( a(t) = 48 - 12t \). This new function tells us how the speed of the train is increasing or decreasing. In essence, if acceleration is positive, velocity is increasing; if negative, velocity is decreasing.

For example, when calculating at \( t = 1 \) hour, we see that:

• Acceleration \( a(1) = 36 \) miles per hour squared.

This positive acceleration indicates the train was speeding up at that point in time. With the relationship of velocity and acceleration, we not only watch how fast something moves but also how it gets faster or slower, providing a fuller picture of its movement dynamics.

Derivative

Derivatives are central in differential calculus and help us analyze changes. They act as the mathematical technique to compute the rate at which one quantity changes with respect to another. Key to this exercise, derivatives help us find velocity from position functions and acceleration from velocity functions.

Let's break this down:

• Deriving the position function \( s(t) = 24t^2 - 2t^3 \) gives us the velocity, \( v(t) = 48t - 6t^2 \). This transformation allows us to translate position into a dynamic understanding of motion.
• Continuing this process by deriving the velocity yields the acceleration, \( a(t) = 48 - 12t \). Here, derivatives unveil how the change rate of speed itself is changing.

Using derivatives helps turn static data (positions) into dynamic insights (speeds and acceleration). For students, understanding this transformation provides not only the practical steps in solving equations but also a foundation for deeper learning in physics and applied mathematics.