A person walks first at a constant speed of 5.10 m/s along a straight line from point circle a to point circle b and then back along the line from circle b to circle a at a constant speed of 2.95 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

Respuesta :

Answer:

(A) Average speed is 3.738 m/sec

(B) Average velocity is 0 m/sec

Explanation:

Let the distance from point a and b is d

Speed of person in going from a to b = 5.10 m/sec

So time taken to reach b from a [tex]=\frac{d}{5.10}sec[/tex]

Ans speed in coming back from b to a is 2.95 m/sec

So time taken in coming from b to a [tex]=\frac{d}{2.95}sec[/tex]

(A) Total distance = d+d = 2d

Total time = [tex]t=\frac{d}{5.10}+\frac{d}{2.95}=0.1960d+0.338d=0.5349d[/tex]

Average speed is equal to ratio of total distance and total time

So average speed [tex]=\frac{2d}{0.5349d}=3.738m/sec[/tex]

(B) As the person goes from a to b and then come back to a from b

So displacement = d-d = 0

So average velocity [tex]=\frac{0}{0.5349d}=0m/sec[/tex]

Otras preguntas

(3) рдХрдерд╛рд▓реЗрдЦрди : (80 рддреЗ 90 рд╢рдмреНрдж) рдкреБрдвреАрд▓ рдореБрджреНрджреЗ рд▓рдХреНрд╖рд╛рдд рдШреЗрдКрди рд╕рдорд░реНрдкрдХ рдХрдерд╛ рдереЛрдбрдХреНрдпрд╛рдд рд▓рд┐рд╣рд╛ : рдЧрд░реАрдм рдореБрд▓рдЧрд╛ рдорд┐рдард╛рдИрдЪреНрдпрд╛ рджреБрдХрд╛рдирд╛рдкреБрдвреЗ рдорд┐рдард╛рдИрд╡рд╛рд▓рд╛ рд╡рд╛рд╕рд╛рдЪреА рдХрд┐рдВрдордд рдорд╛рдЧрддреЛ рдЪрд╛рдгрд╛рдХреНрд╖ рд╡рдХреАрд▓ рдРрдХрдд