Question

What is the magnitude of an object's average velocity if the object moves from a point with coordinates \(x=2.0 \mathrm{~m}, y=-3.0 \mathrm{~m}\) to a point with coordinates \(x=5.0 \mathrm{~m}, y=-9.0 \mathrm{~m}\) in a time interval of \(2.4 \mathrm{~s} ?\)

Short answer

Answer: The magnitude of the average velocity is \(\frac{3\sqrt{5}}{2.4}\, \mathrm{m/s}\).

Step-by-step solution

Find the Displacement in the x-direction

To find the displacement in the x-direction, we will subtract the initial x-coordinate from the final x-coordinate: \((5.0 - 2.0) \mathrm{~ m} = 3.0\, \mathrm{m}\).

Find the Displacement in the y-direction

To find the displacement in the y-direction, we will subtract the initial y-coordinate from the final y-coordinate: \((-9.0 - (-3.0))\mathrm{~ m} = -6.0\, \mathrm{m}\).

Find the Magnitude of the Displacement Vector

Use the Pythagorean theorem to find the magnitude of the displacement vector: \(\sqrt{(3.0\, \mathrm{m})^2 + (-6.0\, \mathrm{m})^2} = \sqrt{45}\, \mathrm{m} = 3\sqrt{5}\, \mathrm{m}\).

Find the Magnitude of the Average Velocity

Divide the magnitude of the displacement by the time interval to find the magnitude of the average velocity. \((3\sqrt{5}\, \mathrm{m}) / (2.4\, \mathrm{s}) = \frac{3\sqrt{5}}{2.4}\, \mathrm{m/s}\).

Key concepts

Displacement Vector

Understanding the displacement vector is crucial when studying motion in physics. It signifies the change in position of an object and is a vector quantity, which means it has both magnitude and direction. To calculate the displacement vector, you need to know the object's initial and final positions.

In the given exercise, the object moved from one point to another, described by the change from an initial set of coordinates \( (2.0, -3.0) \) to a final set \( (5.0, -9.0) \). To find the displacement vector, we subtract the initial coordinates from the final coordinates for both the x and y components. This calculation reveals how far and in what direction the object has moved in each dimension, forming the components of the displacement vector.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry, often used when dealing with right-angled triangles. It states that in a right triangle, the square of the length of the hypotenuse \(c\text{-side})\) is equal to the sum of the squares of the other two sides \(a\text{-side and }b\text{-side})\). Written as an equation, it is \(c^2 = a^2 + b^2\).

In our case, the displacements along the x and y axes form the two legs of a right-angled triangle, while the displacement vector acts as the hypotenuse. The lengths of the x and y displacement become \(a\) and \(b\), respectively. We use the Pythagorean theorem to find the magnitude of the displacement vector, effectively calculating the 'straight-line' distance the object traveled between the two points.

Coordinates

Coordinates are a set of values that show an exact position. In a two-dimensional space, such as a plane, these values are usually denoted as \(x\) and \(y\text{-coordinates})\). The \(x\text{-coordinate})\) shows the position along the horizontal axis, while the \(y\text{-coordinate})\) shows the position along the vertical axis.

For motion problems, these coordinates help determine the location of points in space. By comparing the initial and final coordinates of an object, like we did in our exercise, we can track its movement and assess its behavior through such vectors as displacement.

Magnitude Calculation

The magnitude of a vector is a measure of its 'size' or 'length'. To compute the magnitude of a displacement vector, we combine the components of displacement calculated from the coordinates using the Pythagorean theorem, as these components represent the lengths of the sides of the right triangle that the displacement vector forms.

After finding the magnitude of the displacement in our exercise, the next step is to determine the magnitude of the average velocity. This is achieved by dividing the magnitude of the displacement by the time interval during which the displacement occurred, resulting in a scalar value, meaning it only has magnitude and no direction, measured in units of distance per time (such as meters per second).