Question

A car travels along a straight line at a constant speed of 55.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 26.0 mi/h.
(a) What is the constant speed with which the car moved during the second distance d?
(b) Suppose the second distance d were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip?
(c) What is the average speed for this new trip?

Answer 1

Answer:

a. 17.03 mi/hr b. 0 mi/hr or 18.99 mi/hr c. 17.03 mi/hr

Step-by-step explanation:

Av Vel =(V1+V2)/2 =26

Av Speed =d/t=55=> d=55t1

D=d+d 2d =2(55t1)=110t1

av Vel =110ti/(t1+t2)

t2=3.23t1 => total time =4.23 t1

a. V2=55t1/3.23t1 =17.03 mi/hr

b. velocity as vector quantity (v2-v1)/2 =18.88 or displacement d-d=0 and av velocity = (d-d)/2= 0/2

c. since all information are standing but opposite direction which doesn't affect speed as scalar av speed is the same