A car travels along a straight line at a constant speed of 49.0 mi/h for a dista
ID: 1295253 • Letter: A
Question
A car travels along a straight line at a constant speed of 49.0 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 34.5 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?
mi/h
(c) What is the average speed for this new trip?
mi/h
Explanation / Answer
Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.
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?
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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