Question

A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 2.90 m/s. What is the average speed over the entire trip?

Answer 1

This problem is a very interesting problem.
we know wacth:

Speed = distance / time.

Consider that:
distance from point A to point B=distance from point B to point A=d

We calculate the time to go from point A to point B.
time=distance / speed.
T₁=d / (5 m/s)

We calculate the time to go from point B to point A.
T₂=d / (2.9 m/s)

Therefore; the total time wil be: T₁+T₂
Total time=d/(5 m/s) + d / (2.9 m/s)
Least common multiple=(5 m/s)(2.9 m / s)=14.5 m²/s²
Total time=(2.9 m/s d + 5 m/s d)/ 14.5m²/s²
Total time=[(7.9 d) m/s] / (14.5 m²/s²)
Total time=7.9 d/ (14.5 m/s)

Now, the total distance will be =d+d=2d

Therefore:
Average speed=total distance / total time
Average speed=[(2d)]   /  [(7.9 d)/ (14.5 m/s)]
Average speed≈3.67 m/s.


Answer: the average speed will be ≈3.67 m/s.

Answer 2

The Average Speed over the entire trip is 3.671 meters per second.

The person is moving in a straight line at a constant Speed. Let suppose that Distance between points A and B is $r$, then we construct and use the Kinematic formulas for each stage of the travel to determine Times and Average Speed, based on the fact that Speed is inversely proportional to Time, we derive the resulting expression:

First Stage (from point A to point B):

$t_{1} = \frac{r}{v_{1}}$ (1)

Second Stage (from point B to point A):

$t_{2} = \frac{r}{v_{2}}$ (2)

Average Speed:

$t_{1}+t_{2} = \frac{2\cdot r}{\bar v}$ (3)

Where:

$t_{1}$ - Travelling time for the first stage, in seconds.

$t_{2}$ - Travelling time for the second stage, in seconds.

$r$ - Distance from A to B, in meters.

$v_{1}$ - Speed of the person in the first stage, in meters per second.

$v_{2}$ - Speed of the person in the second stage, in meters per second.

$\bar v$ - Average speed, in meters per second.

By applying (1) and (2) in (3), we derive an expression to determine the Average Speed:

$\frac{r}{v_{1}} + \frac{r}{v_{2}} = \frac{2\cdot r}{\bar v}$

$\frac{v_{2}+v_{1}}{v_{1}\cdot v_{2}} = \frac{2}{\bar v}$

$\bar v = \frac{2\cdot (v_{1}\cdot v_{2})}{v_{2}+v_{1}}$ (4)

If we know that $v_{1} = 5\,\frac{m}{s}$ and $v_{2} = 2.90\,\frac{m}{s}$, the average speed over the entire trip is:

$\bar v = \frac{2\cdot (v_{1}\cdot v_{2})}{v_{2}+v_{1}}$

$\bar v = \frac{2\cdot \left(5\,\frac{m}{s} \right)\cdot \left(2.90\,\frac{m}{s} \right)}{2.90\,\frac{m}{s} + 5\,\frac{m}{s} }$

$\bar v = 3.671\,\frac{m}{s}$

The Average Speed over the entire trip is 3.671 meters per second.

Please see this question related to Average Speed for further details: https://brainly.com/question/19335778