Question

1. -/2 points a person walks first at a constant speed of 5.30 m/s along a straight line from point ⓐ to point ⓑ and then back along the line from ⓑ to ⓐ at a constant speed of 2.90 m/s. (a) what is her average speed over the entire trip? m/s (b) what is her average velocity over the entire trip? m/s resources read it master it details my notes ask your teacher practice anot

Explanation:

Step1: Define the formula for average speed

Average speed $v_{avg}=\frac{\text{total distance}}{\text{total time}}$. Let the distance from A to B be $d$. The total distance $D = d + d=2d$. The time taken from A to B, $t_1=\frac{d}{v_1}$ where $v_1 = 5.30$ m/s, and the time taken from B to A, $t_2=\frac{d}{v_2}$ where $v_2 = 2.90$ m/s.

Step2: Calculate the total time

$t_{total}=t_1 + t_2=\frac{d}{5.30}+\frac{d}{2.90}=d(\frac{1}{5.30}+\frac{1}{2.90})=d(\frac{2.90 + 5.30}{5.30\times2.90})=d\frac{8.20}{15.37}$.

Step3: Calculate the average speed

$v_{avg}=\frac{2d}{t_{total}}=\frac{2d}{d\frac{8.20}{15.37}}=\frac{2\times15.37}{8.20}\approx3.75$ m/s.

Step4: Define the formula for average velocity

Average velocity $\vec{v}_{avg}=\frac{\text{total displacement}}{\text{total time}}$. Since the person returns to the starting - point, the total displacement $\Delta\vec{x}=0$.

Step5: Calculate the average velocity

$\vec{v}_{avg}=\frac{0}{t_{total}} = 0$ m/s.

Answer:

(a) 3.75 m/s
(b) 0 m/s