A car travels along a straight line at a constant speed of 51.5 mi/h for a dista

Question

A car travels along a straight line at a constant speed of 51.5 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 29.5 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?

Explanation / Answer

(a) assume the speed for the 1st distance and 2nd distance are v1 and v2 respectively

the time used for 1st distance  t1 = d/v1

time used for the 2nd distance  t2 = d/v2

total distance is d+d = 2d

avg speed = 2d/(t1+t2) = 2d /(d/v1+ d/v2) = 2 v1v2/(v1+v2)

plug in the respective values

29.5 = 2*51.5*v2/(51.5+v2)

29.5*(51.5+v2) = 2*51.5*v2

v2 = 29.5*51/(2*51.5-29.5) = 20.5 mi/h

(b) since the net displacement is zero, the average velocity = displacement/time = 0

(c) the speed = distance/time, it is same as 29.5 mi/h

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