Problem 87 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

DISTANCE AND VELOCITY After \(t\) minutes, an object moving along a line has velocity \(v(t)=1+4 t+3 t^{2}\) meters per minute. How far does the object travel during the third minute?

Step-by-Step Solution

Answer

The object travels 30 meters during the third minute.

Step 1: Understand the Problem

Determine the distance traveled by the object during the third minute. The given velocity function is \(v(t)=1+4 t+3 t^{2}\) meters per minute.

Step 2: Determine the Time Interval for the Third Minute

The third minute corresponds to the time interval from \(t=2\) to \(t=3\) minutes.

Step 3: Set Up the Integral for Distance

The distance traveled is the integral of the velocity function over the given interval. Set up the integral for the velocity function from \(t=2\) to \(t=3\): \[\text{Distance} = \int_{2}^{3} (1 + 4t + 3t^{2}) \, dt\]

Step 4: Evaluate the Integral

Compute the integral by finding the antiderivative of the velocity function and then evaluating it from 2 to 3. The antiderivative of \(1+4t+3t^2\) is: \[\text{Antiderivative} = t + 2t^2 + t^3\]Evaluate it at the boundaries: \[\text{Distance} = \left[t + 2t^2 + t^3 \right]_{2}^{3}\]

Step 5: Calculate the Result

Substitute the boundary values into the antiderivative: \[\text{At } t=3: \left(3 + 2(3)^2 + (3)^3 \right) = 3 + 2 \cdot 9 + 27 = 3 + 18 + 27 = 48\]\[\text{At } t=2: \left(2 + 2(2)^2 + (2)^3 \right) = 2 + 2 \cdot 4 + 8 = 2 + 8 + 8 = 18\]Now, find the difference: \[\text{Distance} = 48 - 18 = 30 \text{ meters}\]

Key Concepts

definite integral

In calculus, a definite integral represents the area under a curve within a specified interval. For this problem, it shows the total distance traveled by an object when its velocity changes over time. To find the distance traveled during the third minute, we compute the definite integral of the velocity function between 2 and 3 minutes. We set up the integral as \(\text{Distance} = \int_{2}^{3} (1 + 4t + 3t^{2}) \, dt\). This operation captures all velocity changes within that minute and aggregates them to calculate the total distance.

velocity function

The velocity function describes how an object’s speed changes over time. For this problem, the velocity function is given by \(v(t) = 1 + 4t + 3t^2\), where *t* is time in minutes, and the result is in meters per minute. By plugging in different values of *t*, we can find the exact velocity at any moment. For instance, at \(t = 2\), the velocity is \(v(2) = 1 + 4(2) + 3(2^2) = 1 + 8 + 12 = 21\) meters per minute. As *t* increases, the velocity changes accordingly, following the pattern described by the equation.

antiderivative

The antiderivative of a function is a new function whose derivative results in the original function. In this problem, we need the antiderivative of the velocity function \(1 + 4t + 3t^2\). The antiderivative is calculated by integrating each term separately:

• The antiderivative of 1 is \(t\)
• The antiderivative of \(4t\) is \(2t^2\)
• The antiderivative of \(3t^2\) is \(t^3\)

So the full antiderivative is \(t + 2t^2 + t^3\). We use this to evaluate the definite integral.

distance traveled

To find the distance traveled during a specific time interval, we evaluate the definite integral of the velocity function. For this problem, we compute the integral from 2 to 3 minutes:

• Evaluate the antiderivative at the upper limit (\t=3): \(3 + 2(3^2) + (3^3) = 3 + 18 + 27 = 48\)
• Evaluate the antiderivative at the lower limit (\t=2): \(2 + 2(2^2) + (2^3) = 2 + 8 + 8 = 18\)

Subtract the lower value from the upper value: \(48 - 18 = 30\). Hence, the distance traveled during the third minute is 30 meters.