Problem 80 - Calculus and its applications

Question

A particle is released during an experiment. Its speed \(t\) minutes after release is given by \(v(t)=-0.3 t^{2}+9 t,\) where \(v(t)\) is in kilometers per minute. a) How far does the particle travel during the first \(10 \mathrm{~min} ?\) b) How far does it travel during the second \(10 \mathrm{~min} ?\)

Step-by-Step Solution

Answer

The particle travels 350 km in the first 10 minutes and 650 km in the second 10 minutes.

Step 1: Understanding the problem

To find out how far the particle travels, we need to calculate the integral of the velocity function over the given time intervals, since the integral of velocity gives us the distance.

Step 2: Find the distance for the first 10 minutes

The distance traveled in the first 10 minutes is obtained by evaluating the definite integral of the velocity function from 0 to 10. Thus, calculate \[\int_{0}^{10} (-0.3t^2 + 9t) \, dt\].

Step 3: Integrate the velocity function

First, compute the antiderivative of the velocity: \[\int (-0.3t^2 + 9t) \, dt = -0.1t^3 + 4.5t^2 + C\] where \(C\) is the constant of integration which will be eliminated when evaluating the definite integral.

Step 4: Evaluate the integral from 0 to 10 minutes

Substitute the limits 0 and 10 into the antiderivative: \[\left[ -0.1(10)^3 + 4.5(10)^2 \right] - \left[ -0.1(0)^3 + 4.5(0)^2 \right]\]Simplifying this gives: \[-0.1(1000) + 4.5(100) = -100 + 450 = 350 \] km.

Step 5: Find the distance for the second 10 minutes

The second 10 minutes cover the interval from \(t=10\) to \(t=20\). Calculate \[\int_{10}^{20} (-0.3t^2 + 9t) \, dt\].

Step 6: Evaluate the integral from 10 to 20 minutes

Using the antiderivative from Step 3, substitute the limits 10 and 20: \[\left[ -0.1(20)^3 + 4.5(20)^2 \right] - \left[ -0.1(10)^3 + 4.5(10)^2 \right]\]Simplifying gives: \(-0.1(8000) + 4.5(400) = -800 + 1800 = 1000 \) km; subtract the previous result to focus on this interval: \(1000 - 350 = 650\) km.

Key Concepts

Definite Integrals

In integral calculus, one of the core concepts is the "definite integral." Definite integrals are used to calculate the net area under a curve, usually representing a function, over a specified interval. They provide a powerful tool in finding collective quantities such as total distance, area, or accumulated change.

The definite integral \(\int_{a}^{b} f(x) \, dx\) involves computing the antiderivative of the function \(f(x)\) and evaluating it at the upper (\