Question

A student walks to the right, 25-m along the C Hall in 15-s. They turn around and walk 15-m to the left in 8.0-s. What is the average speed and the average velocity? (V represents average speed and ΔV represents average velocity)

Answer 1

For the student that walks to the right 25 m in 15 s and then turns and walks to the left 15 m in 8.0 s, we have:

1. The average speed is 1.74 m.

2. The average velocity is 1.42 m.

1. The average speed is an scalar wich is given by:

$V = \frac{d_{t}}{t_{t}}$

Where:

$d_{t}$: is the total distance

$t_{t}$: is the total time

The total distance and the total time can be calculated as follows:

$d_{t} = d_{1} + d_{2} = 25 m + 15 m = 40 m$

$t_{t} = t_{1} + t_{2} = 15 s + 8.0 s = 23 s$

Since V is a scalar, the direction of motion is not important to calculate the final value.

Then, the average speed is:

$V = \frac{d_{t}}{t_{t}} =\frac{40 m}{23 s} = 1.74 m/s$

2. The average velocity is a vector, so the direction of motion is important here. This ΔV can be calculated with the following equation:

$\Delta V = \frac{\Delta x}{\Delta t} = \frac{x_{f} - x_{i}}{t_{f} - t_{i}}$

Where:

x: is the displacement  

f: is for final and i: is for initial

Hence, the average velocity is:

$\Delta V = \frac{x_{f} - x_{i}}{t_{f} - t_{i}} = \frac{15 m - 25 m}{8.0 s - 15 s} = 1.42 m/s$

You can find more about average speed and average velocity here: https://brainly.com/question/12322912?referrer=searchResults

I hope it helps you!