Question

Find the displacement, distance traveled, average velocity and average speed of the described object on the given interval. An object with velocity function \(\vec{v}(t)=\langle 1,2,-1\rangle,\) where distances are measured in feet and time is in seconds, on [0,10]

Short answer

Displacement: \(\langle 10, 20, -10 \rangle\); Distance: \(10\sqrt{6}\) feet; Average Velocity: \(\langle 1, 2, -1 \rangle\) ft/s; Average Speed: \(\sqrt{6}\) ft/s.

Step-by-step solution

Understanding the Velocity Function

The given velocity function is constant: \(\vec{v}(t)=\langle 1,2,-1\rangle\). This indicates that the object moves with a constant velocity of 1 ft/s in the x-direction, 2 ft/s in the y-direction, and -1 ft/s in the z-direction.

Find Displacement

The displacement is found by integrating the velocity function over the given time interval. Since the velocity is constant, displacement \(\vec{d}\) can be calculated as:\[\vec{d} = \int_0^{10} \vec{v}(t) \, dt = \vec{v}(t) \times 10 = \langle 1 \times 10, 2 \times 10, -1 \times 10 \rangle = \langle 10, 20, -10 \rangle\]The displacement vector is \(\langle 10, 20, -10 \rangle\) feet.

Calculate Distance Traveled

The distance traveled is the integral of the magnitude of the velocity vector over the time interval [0,10]. Since the velocity vector is constant, the magnitude of the velocity \(||\vec{v}||\) is\[\sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} \, \text{ft/s}\]Then, the distance traveled is:\[\text{Distance} = \int_0^{10} ||\vec{v}|| \, dt = ||\vec{v}|| \times 10 = \sqrt{6} \times 10 = 10\sqrt{6}\,\text{feet}\]

Calculate Average Velocity

The average velocity \(\vec{v}_{avg}\) over the interval is the total displacement divided by the total time:\[\vec{v}_{avg} = \frac{\vec{d}}{10} = \frac{\langle 10, 20, -10 \rangle}{10} = \langle 1, 2, -1 \rangle \, \text{ft/s}\]Since the velocity is constant, the average velocity is the same as the velocity vector.

Calculate Average Speed

The average speed is the total distance traveled divided by the total time:\[\text{Average Speed} = \frac{\text{Distance}}{10} = \frac{10\sqrt{6}}{10} = \sqrt{6} \, \text{ft/s}\]The average speed is \(\sqrt{6}\,\text{ft/s}\).

Key concepts

Displacement

In vector calculus, displacement refers to the change in position of an object, described as a vector. It is a crucial concept when studying motion. Displacement considers only the initial and final positions, not the path taken.
In our problem, we deal with a constant velocity vector of \(\vec{v}(t)=\langle 1,2,-1\rangle\). Displacement \(\vec{d}\) can be calculated by integrating the velocity function over time. However, since the velocity is constant, it's simpler to multiply the velocity by the time interval:
You'll find:

• \(\vec{d} = \int_0^{10} \vec{v}(t) \, dt = \vec{v}(t) \times 10 = \langle 1 \times 10, 2 \times 10, -1 \times 10 \rangle = \langle 10, 20, -10 \rangle\)

The resulting displacement vector \(\langle 10, 20, -10 \rangle\) indicates the net change in position over the 10-second interval.

Velocity

Velocity is a vector quantity that describes the rate of change of an object's position. It has both magnitude and direction, distinguishing it from speed, which is scalar.
In our exercise, the velocity vector is \(\vec{v}(t)=\langle 1,2,-1\rangle\). This means:

• 1 ft/s in the x-direction
• 2 ft/s in the y-direction
• -1 ft/s in the z-direction

The negative sign in the z-direction suggests movement opposite to initial orientation.
Constant velocity implies no acceleration, so the motion is straightforward. This simplification makes calculations simpler, as the velocity doesn’t change over time.

Distance Traveled

Distance traveled accounts for the entire path an object follows, as opposed to only the change in position. When dealing with vector velocity, it’s calculated by integrating the magnitude of velocity over the given time period.
For the given problem, the magnitude of velocity \(||\vec{v}||\) is calculated as follows:
\[||\vec{v}|| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}\]
The complete distance traveled over the 10 seconds is:

• \(10\sqrt{6}\) feet.

This measure highlights how much ground the object has covered, regardless of starting and ending point comparison.

Average Velocity

Average velocity provides a summarization of an object's entire movement over a set period of time. It's discovered by dividing the total displacement by the total duration.
For this exercise, the average velocity \(\vec{v}_{avg}\) can be computed with:
\[\vec{v}_{avg} = \frac{\vec{d}}{10} = \frac{\langle 10, 20, -10 \rangle}{10} = \langle 1, 2, -1 \rangle \text{ ft/s}\]
Here, because the velocity within the interval is constant, the average velocity is identical to the standard velocity vector. This statement aligns with how average velocity often simplifies complicated movement into a cleaner model, pinpointing directions and rates during specified periods.