Problem 62 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that an object, moving along the \(x\) -axis, has velocity \(v=t^{2}\) meters per second at time \(t\) seconds. How far did it travel between \(t=3\) and \(t=5\) ? See Problem 61 .

Step-by-Step Solution

Answer

The object travels approximately 32.67 meters.

Step 1: Recognize the problem

The problem requires finding the total distance traveled by an object with a given velocity function between two time intervals. The velocity function given is \( v(t) = t^2 \) m/s and we are asked to find the distance traveled between \( t = 3 \) s and \( t = 5 \) s.

Step 2: Set up the integral for distance

The distance traveled by an object can be found by integrating the velocity function over the given time interval. This means we need to calculate:\[\int_{3}^{5} v(t) \ dt = \int_{3}^{5} t^2 \ dt\].

Step 3: Compute the integral

To solve the integral,\[ \int t^2 \, dt = \frac{t^3}{3} + C \].Now, evaluate this from \( t = 3 \) to \( t = 5 \).

Step 4: Evaluate the definite integral

Substitute the integration limits into the antiderivative:\[\left[ \frac{t^3}{3} \right]_{3}^{5} = \frac{5^3}{3} - \frac{3^3}{3}\].Calculate each term:- \( \frac{5^3}{3} = \frac{125}{3} \)- \( \frac{3^3}{3} = \frac{27}{3} = 9 \).

Step 5: Calculate the distance

Subtract the two values obtained in the previous step:\[ \frac{125}{3} - 9 = \frac{125}{3} - \frac{27}{3} = \frac{98}{3} \].Convert this to a decimal if necessary.\[ \frac{98}{3} \approx 32.67 \] meters.

Key Concepts

Velocity Function

A velocity function describes how the velocity of an object changes with time. In the context of motion along a straight line, the velocity function tells us the speed and direction at any given moment. For example, in our problem, the velocity function is given by \( v(t) = t^2 \), which indicates that the velocity changes as a square function of time. This means as time increases, the velocity increases quadratically.

• A positive velocity implies that the object is moving in the positive direction along the x-axis.
• If the velocity were negative, it would indicate motion in the opposite direction

Understanding the velocity function is crucial for determining how far an object travels over time, especially when you need to calculate the total distance or displacement.

Definite Integral

The definite integral is a powerful tool in calculus that helps determine the accumulation of quantities, such as distance, area, or volume. It provides the total amount of change in a quantity, given its rate of change, over a specific interval. In this problem, the integral of the velocity function over time reveals the distance traveled. When we set up the integral \( \int_{3}^{5} t^2 \, dt \), we are effectively summing up all the small increments of distance over the interval from \( t=3 \) seconds to \( t=5 \) seconds. The process involves finding the antiderivative of the function \( t^2 \) and then evaluating it at the boundaries.

• Antiderivative: The antiderivative of \( t^2 \) is \( \frac{t^3}{3} + C \).
• Evaluation: Substitute the limits of integration into the antiderivative and compute the difference.

Through this process, we translate the velocity data into a meaningful measurement of distance traveled.

Distance Traveled

Distance traveled refers to the total path covered by an object, regardless of its direction or changes in direction. It is always a non-negative value. In problems involving a velocity function, integrating this function over a given time interval gives us the distance traveled. Calculating the definite integral of \( v(t) = t^2 \) from \( t=3 \) to \( t=5 \) provides:1. Step by step application of the Fundamental Theorem of Calculus.2. Finding \[\int_{3}^{5} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{3}^{5} = \frac{125}{3} - 9 = \frac{98}{3} \approx 32.67 \text{ meters}\]

• This result tells us that over the interval from 3 to 5 seconds, the object has covered approximately 32.67 meters.
• It's important to note that this value represents the total distance traveled, rather than a net displacement from the starting point.

Understanding the distance traveled is key in many physics and engineering applications, where precise measurements ensure accuracy and efficiency.