Question

A person walks first at a constant speed of 5 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 m / s.

a. What is her average speed over the entire trip? b. Her average velocity over the entire trip?

Please answer it's very urgent and please don't scam

Answer 1

Answer:

Speed is positive whenever motion occurs, so the average speed must be positive. For the velocity, we take as positive for motion to the right and negative for motion to the left, so its average value can be positive, negative, or zero.    

(a) The average speed during any time interval is equal to the total distance of travel divided by the total time:

Average speed = total distance / total time

=tAB+tBAdAB+dBA

But dAB=dBA,tAB=d/VAB, and tBA=d/VBA

So, average speed = (d/VAB)+(d/VBA)d+d=VAB+VBA2(VAB)(VBA)

and

average speed = 2[5.00m/s+3.00m/s(5.00m/s)(3.00m/s)]=3.75m/s.

(b) The average velocity during any time interval equals total displacement divided by elapsed time

Vx,avg=ΔtΔx

Since the walker returns to the starting point, Δx=0 and Vx,avg=0.

Explanation:

Hope it helps.

put average speed as 2 according to your question

Answer 2

Answer: Average speed is 1.42 m/s and the average velocity is 0 m/s.

Given Information: A person walks first at a constant speed of 5 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 m/s.

To Calculate: Average speed and average velocity.

Explication of steps: As per the given question, we have :

• Speed from A to B = 5 m/s
• Speed from B to C = 2 m/s

Let us suppose the distance from A to B as x m. In order to calculate the average velocity, we need to calculate the total distance and total time first. Let's find out total distance:

➝ Total distance = Distance (A to B) + Distance (B to C)

➝ Total distance = (x + x) m

➝ Total distance = 2x m

Now, finding out total time taken.

➝ Total time = Time(A to B) + Time(B to C)

• Time = Distance/Speed

➝ Total time = $\sf \dfrac{x}{5} + \dfrac{x}{2}$ s

➝ Total time = $\sf \dfrac{2x+ 5x}{5}$ s

➝ Total time = $\sf \dfrac{7x}{5}$ s

Now,as we know that :

$\large{\underline{\boxed { \sf {Speed_{(Avg)} = \dfrac{Distance_{(Total)}}{Time_{(Total)}} }}}}$

$\leadsto\sf { Speed_{(Avg)} = \Bigg [ 2x \div \dfrac{7x}{5} \Bigg ] \; ms^{-1}}$

$\leadsto\sf { Speed_{(Avg)} = \Bigg [ 2x \times \dfrac{5}{7x} \Bigg ] \; ms^{-1}}$

$\leadsto\sf { Speed_{(Avg)} = \Bigg [ \cancel{\dfrac{10x}{7x} }\Bigg ] \; ms^{-1}}$

$\leadsto\sf { Speed_{(Avg)} = 1.42 \; ms^{-1}}$

Therefore, average speed of the person is 1.42 m/s.

Now, we are asked to calculate the average velocity of the person.

$\large{\underline{\boxed { \sf {Velocity_{(Avg)} = \dfrac{Displacement_{(Total)}}{Time_{(Total)}} }}}}$

Here, the displacement is 0 as the body comes back to its initial position (A) after covering certain distance. This means initial position and final position is same, thus displacement is 0. Total time is 7x/5 s.

➝ Average velocity = (0 ÷ 7x/5) m/s

➝ Average velocity = 0 m/s

Therefore, average velocity of the person is 0 m/s.