Problem 59 - Calculus: Early Transcendentals

Question

The velocity function (in meters per second) is given for a particle moving along a line. Find (a) the displacement and (b) the distance traveled by the particle during the given time interval. \( v(t) = 3t - 5 \), \( 0 \le t \le 3 \)

Step-by-Step Solution

Answer

(a) Displacement is -1.5 meters. (b) Distance is 5.8333 meters.

Step 1: Understanding the Problem

We need to find two things: (a) the displacement of the particle and (b) the distance traveled by the particle over the time interval from \( t = 0 \) to \( t = 3 \). We are given the velocity function \( v(t) = 3t - 5 \).

Step 2: Finding Displacement

Displacement is the net change in position of the particle, calculated using the definite integral of the velocity function over the given interval. We compute \[\text{Displacement} = \int_{0}^{3} (3t - 5) \, dt.\]

Step 3: Integrating the Velocity Function

Calculate the antiderivative of \( 3t - 5 \):\[\int (3t - 5) \, dt = \frac{3}{2}t^2 - 5t + C.\]

Step 4: Evaluating the Definite Integral for Displacement

To find the displacement, evaluate the definite integral:\[\int_{0}^{3} (3t - 5) \, dt = \left[ \frac{3}{2}t^2 - 5t \right]_{0}^{3}.\]Calculate:\[= \left(\frac{3}{2}(3)^2 - 5(3)\right) - \left(\frac{3}{2}(0)^2 - 5(0)\right)\]\[= \left(\frac{27}{2} - 15\right) - (0)\]\[= \left(13.5 - 15\right)\]\[= -1.5.\]

Step 5: Understanding Distance Traveled

Distance is the total length of path traveled without considering direction, calculated as the integral of speed, which is the absolute value of velocity: \[\text{Distance} = \int_{0}^{3} |3t - 5| \, dt.\]

Step 6: Determining When Velocity is Zero

Before solving for distance, find when velocity changes direction by setting \( v(t) = 0 \):\[3t - 5 = 0.\]\[3t = 5.\]\[t = \frac{5}{3}.\]The particle changes direction at \( t = \frac{5}{3} \).

Step 7: Calculating the Distance in Two Intervals

Split the integration:1. From \( t = 0 \) to \( t = \frac{5}{3} \): velocity is negative, so use \(|v(t)| = -(3t - 5)\).2. From \( t = \frac{5}{3} \) to \( t = 3 \): velocity is positive, so use \(|v(t)| = 3t - 5\).Evaluate:\[\int_{0}^{5/3} -(3t - 5) \, dt + \int_{5/3}^{3} (3t - 5) \, dt.\]

Step 8: Evaluating the Integral Parts Separately

Calculate the first part:\[\int_{0}^{5/3} -(3t - 5) \, dt = \left[-\left(\frac{3}{2}t^2 - 5t\right)\right]_{0}^{5/3}.\]Calculate:\[= -\left(\frac{3}{2}(\frac{5}{3})^2 - 5(\frac{5}{3})\right) + 0\]\[= -\left(\frac{25}{6} - \frac{25}{3}\right)\]\[= \frac{25}{6}.\]Now, calculate the second part:\[\int_{5/3}^{3} (3t - 5) \, dt = \left[\frac{3}{2}t^2 - 5t\right]_{5/3}^{3}.\]Calculate:\[= \left(\frac{27}{2} - 15\right) - \left(\frac{25}{6} - \frac{25}{3}\right)\]\[= 1.5 + \frac{25}{6}.\]Combine these:\[\text{Total Distance} = \frac{25}{6} + (1.5) = 4.3333 + 1.5 = 5.8333.\]

Step 9: Finalizing the Results

Conclude the results:(a) Displacement = \(-1.5\) meters(b) Distance = \(5.8333\) meters.

Key Concepts

Displacement

Displacement is a key concept in calculus, representing the net change in position of an object along a straight line over a specific time period. To calculate displacement, we integrate the velocity function over the given time interval. This process involves finding the definite integral of the velocity function, which represents the sum of infinitesimal changes in position over time.

For example, with a velocity function given by \( v(t) = 3t - 5 \) and the interval from \( t = 0 \) to \( t = 3 \), the displacement is calculated using:

• Displacement \( = \int_{0}^{3} (3t - 5) \, dt \).

This integral yields a value that may be positive, negative, or zero, indicating the final position relative to the starting point.

Definite Integral

A definite integral is a fundamental concept in calculus, used to compute the accumulation of quantities where the boundaries of the integration are defined. Specifically, in the context of displacement, it helps measure the total change in position of an object over time when the velocity varies.

The definite integral takes the form:

• \( \int_{a}^{b} f(t) \, dt \)

where \( a \) and \( b \) are the limits of integration. In our case, \( \int_{0}^{3} (3t - 5) \, dt \) evaluates the net change in position from time 0 to time 3.

Key Steps in Evaluating

The process involves:

• Finding the antiderivative of the velocity function.
• Substituting the upper limit and the lower limit into the antiderivative.
• Calculating the difference, providing the net displacement.

Distance Traveled

The distance traveled differs from displacement, as it refers to the total length of the path taken by an object, regardless of its starting or ending position. This distance is computed by integrating the absolute value of the velocity function, which ensures all portions of the motion contribute positively.

In our example, due to a change in direction when \( v(t) = 0 \) at \( t = \frac{5}{3} \), the integration is split. The calculations involve two intervals:

• From \( t = 0 \) to \( t = \frac{5}{3} \) where velocity is negative.
• From \( t = \frac{5}{3} \) to \( t = 3 \) where velocity is positive.

Adding these results gives the total distance traveled as \( 5.8333 \) meters.

Velocity Function

The velocity function \( v(t) \) provides the rate of change of position with respect to time. It is a derivative of the position function \( s(t) \), capturing how fast and in which direction an object is moving at any time \( t \).

For our problem, \( v(t) = 3t - 5 \) is a linear function, meaning the velocity changes uniformly over time.

Importance of Zero Crossing

When finding the distance traveled, it is crucial to determine when \( v(t) = 0 \) or \( t = \frac{5}{3} \). At this specific time, the particle changes its direction of motion. Knowing this allows us to appropriately split the integral and compute the distance traveled accurately.

By understanding both the direction and magnitude of velocity, we can derive meaningful insights into the particle's overall motion.