Problem 89 - Calculus and its applications

Question

A particle starts at the origin. Its velocity, in miles per hour, after \(t\) hours is given by $$v(t)=3 t^{2}+2 t$$ How far does it travel from the start of the 2 nd hour through the end of the 5 th hour (from \(t=1\) to \(t=5\) )?

Step-by-Step Solution

Answer

The particle travels 148 miles from the start of the 2nd hour to the end of the 5th hour.

Step 1: Understand the Problem

We need to find the distance the particle travels from the start of the 2nd hour to the end of the 5th hour. This translates to finding the distance traveled from time \( t = 1 \) to \( t = 5 \). The distance can be found by integrating the velocity function \( v(t) = 3t^2 + 2t \) over this interval.

Step 2: Set Up the Integral

To find the distance traveled, set up the integral of the velocity function over the interval from \( t = 1 \) to \( t = 5 \). This integral will be \[ \int_{1}^{5} (3t^2 + 2t) \, dt \].

Step 3: Integrate the Function

Calculate the integral \( \int (3t^2 + 2t) \, dt \).The integral of \( 3t^2 \) is \( t^3 \).The integral of \( 2t \) is \( t^2 \).Thus, \( \int (3t^2 + 2t) \, dt = t^3 + t^2 \).

Step 4: Evaluate the Definite Integral

Evaluate the definite integral \[ \int_{1}^{5} (3t^2 + 2t) \, dt \].Using the antiderivative found, compute: \( \left[ t^3 + t^2 \right]_1^5 = (5^3 + 5^2) - (1^3 + 1^2) \).Calculate: \( (125 + 25) - (1 + 1) = 150 - 2 = 148 \).

Step 5: Conclude the Calculation

The integral computation shows the total distance traveled by the particle from \( t=1 \) to \( t=5 \) is 148 miles. This is the solution to the problem.

Key Concepts

Velocity Function

In integral calculus, a velocity function is a mathematical expression used to describe how the velocity of an object changes over time. Velocity, in simple terms, is the rate at which an object moves in a specific direction. When dealing with velocity as a function, you typically see it represented as \( v(t) \), where \( t \) stands for time.
In this particular exercise, we see the velocity function given by \( v(t) = 3t^2 + 2t \). This equation tells us that the particle's velocity is dependent on time \( t \). The function includes both a quadratic term, \( 3t^2 \), and a linear term, \( 2t \).

• The quadratic term (\( 3t^2 \)) suggests that the velocity increases at a rate proportional to the square of time.
• The linear term (\( 2t \)) indicates that the velocity also changes linearly with time.

Understanding this helps us use calculus to find other aspects of the particle's motion, like the distance it travels.

Definite Integral

A definite integral is an important concept in calculus that allows us to calculate the accumulation of quantities, like area under a curve, or in this case, distance traveled over a time interval. When you see the notation \( \int_{a}^{b} f(t) \, dt \), it represents the definite integral of the function \( f(t) \) from \( t = a \) to \( t = b \).
For the given exercise, the definite integral \( \int_{1}^{5} (3t^2 + 2t) \, dt \) is used to find the total distance traveled by the particle from hour 2 to hour 5.

• Setting up the integral establishes the bounds of the problem, from \( t = 1 \) to \( t = 5 \).
• Calculating the integral involves finding an antiderivative for the velocity function.

The process of integrating gives us a function that represents accumulated change (like distance) over the given interval.

Distance Traveled

Distance traveled refers to the total length covered by an object in motion over a specific time period. In the context of the exercise, we use calculus to determine how far the particle has moved from hour 2 to hour 5. This is calculated by evaluating the definite integral of the velocity function over the given time interval.
From the original solution, we integrated the velocity function and then evaluated it as a definite integral:- The antiderivative of \( 3t^2 + 2t \) gives us \( t^3 + t^2 \).- By computing \( \left[ t^3 + t^2 \right]_1^5 = (5^3 + 5^2) - (1^3 + 1^2) \), we determine the distance.
The results show that the particle traveled a total distance of 148 miles from time \( t=1 \) to \( t=5 \). This is a practical application of finding the total distance by integrating a velocity function over a specified interval.