Question

What is the magnitude of an object's average velocity if an 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 object's average velocity is approximately 1.85 m/s.

Step-by-step solution

Calculate the displacement vector between the two points

The displacement vector is the difference between the final and initial positions. Given the initial position coordinates ($x_1=2.0\mathrm{m},y_1=-3.0\mathrm{m}$) and final position coordinates ($x_2=5.0\mathrm{m},y_2=-9.0\mathrm{m}$), the displacement vector $\overrightarrow{d}$ can be calculated as follows: $\overrightarrow{d}=\left(\begin{array}{c}{x}_2-x_1\\ y_2-y_1\end{array}\right)=\left(\begin{array}{c}5.0-2.0\\ -9.0-(-3.0)\end{array}\right)=\left(\begin{array}{c}3.0\\ -6.0\end{array}\right)\mathrm{m}$

Calculate the magnitude of the displacement vector

Now we need to find the magnitude of the displacement vector. For a 2D vector $\overrightarrow{d}=\left(\begin{array}{c}x\\ y\end{array}\right)$, the magnitude $|\overrightarrow{d}|$ is given by the formula: $|\overrightarrow{d}|=\sqrt{x^2+y^2}$ Applying the formula to our displacement vector, we get: $|\overrightarrow{d}|=\sqrt{(3.0{)}^2+(-6.0{)}^2}=\sqrt{9+36}=\sqrt{45}=3\sqrt{5}\mathrm{m}$

Calculate the magnitude of the average velocity

To find the magnitude of the average velocity $v_{avg}$, we will divide the magnitude of the displacement by the time interval $\mathrm{\Delta }t=2.4\mathrm{s}$. Therefore, the magnitude of the average velocity is: $v_{avg}=\frac{|\overrightarrow{d}|}{\mathrm{\Delta }t}=\frac{3\sqrt{5}\mathrm{m}}{2.4\mathrm{s}}\approx 1.85\mathrm{m}/\mathrm{s}$ So, the magnitude of the object's average velocity is approximately $1.85\mathrm{m}/\mathrm{s}$.

Key concepts

Displacement Vector

Understanding what a displacement vector is can make many physics problems much clearer. Simply put, a displacement vector represents the change in position of an object. It is a vector that points from the start position to the end position.
In our example, the object moves from an initial point at coordinates $x_1=2.0\mathrm{m},y_1=-3.0\mathrm{m}$to a final point at $x_2=5.0\mathrm{m},y_2=-9.0\mathrm{m}$.
This move can be visualized with a vector pointing towards the new position. * We calculate the displacement vector, denoted as $\overrightarrow{d}$, by finding the difference in the x-coordinates and y-coordinates: $$\overrightarrow{d}=\left(\begin{array}{c}{x}_2-x_1{y}_2-y_1\end{array}\right)=\left(\begin{array}{c}5.0-2.0-9.0-(-3.0)\end{array}\right)=\left(\begin{array}{c}3.0-6.0\end{array}\right)\mathrm{m}$$
This mathematical tool tells us not only how far the object has moved, but also in which direction.

Magnitude Calculation

When working with vectors, determining their magnitude is essential for understanding how much movement occurred, irrespective of its direction. This is like finding the straight-line distance covered by the object. For this, we use the Euclidean formula in 2D space. * The magnitude of a vector $\overrightarrow{d}=\left(\begin{array}{c}xy\end{array}\right)$ is calculated using:$$|\overrightarrow{d}|=\sqrt{x^2+y^2}$$ In the problem, our displacement vector was $\left(\begin{array}{c}3.0-6.0\end{array}\right)$. So, we calculate:$$|\overrightarrow{d}|=\sqrt{(3.0{)}^2+(-6.0{)}^2}=\sqrt{9+36}=\sqrt{45}=3\sqrt{5}\mathrm{m}$$
This value represents the total distance moved in a straight line, from the start to the finish position. It's critical when comparing the speed of motion over a given interval.

Time Interval

The time interval is a key element when calculating average velocity. It measures how long the object takes to move from its initial position to its final position. Hence, it directly affects the speed measurement.
A time interval is simply the difference between the start and end time of an action. In our scenario, the time interval was given as $\mathrm{\Delta }t=2.4\mathrm{s}$.
* The average velocity is found by dividing the magnitude of the displacement vector by this time interval:$$v_{avg}=\frac{|\overrightarrow{d}|}{\mathrm{\Delta }t}=\frac{3\sqrt{5}\mathrm{m}}{2.4\mathrm{s}}\approx 1.85\mathrm{m}/\mathrm{s}$$ This outcome tells us how fast the object was moving on average during its journey. Average velocity gives insight into standard speeds, smoothing out any fluctuations in actual speed during travel.