Question

A student walks 144 m west, and then turns around and walks 89 m east If this takes place in a 7.5-minute interval, what is the student's average velocity in meters
second? What is the student's average speed?​

Answer 1

The average speed is 0.517 m/s

The average velocity is 0.12 m/s west

Explanation:

When dividing the total distance covered by object by time, we get the value for average speed.

$\text { speed }=\frac{\text {distance}}{\text {time}}$

Calculate the distance without consideration of motion’s direction. So, the distance walks 144 m first, and then another 89 m, so the total distance covered is

d = 144 + 89 = 233 meters

Given t = 7.5 minutes. Convert minute into seconds,  

$7.5 \times 60=450 \text { seconds }$

So, the average speed can be calculates as below,

$\text { speed }=\frac{233}{450}=0.517 \mathrm{m} / \mathrm{s}$

When dividing the object’s displacement by time taken, we can calculate average velocity.

$\text {velocity}=\frac{\text {displacement}}{\text {time}}$

In given case, the students walks 144 m west first, and then 89 m east. The displacement is the distance in a straight line between the initial and final position: therefore, in this case, the displacement is

d = 144 (west) - 89 (east) = 55 m (west)

The time taken is t = 450 s

So, the average velocity is

$\text {velocity}=\frac{55}{450}=0.12 \mathrm{m} / \mathrm{s}$

And the direction is west (the same as the displacement).