A car travels along a straight line at a constant speed of 52.5 mi/h for a dista

Question

A car travels along a straight line at a constant speed of 52.5 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 26.5 mi/h. (a) What is the constant speed with which the car moved during the second distance d? mi/h (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? (Enter the magnitude.) mi/h (c) What is the average speed for this new trip? mi/h

Explanation / Answer

As it is equal distance with different speeds

V1=52.5mi/h(speed of car in first part)

V2=?(speed of car in second part)

Average speed of the trip=26.5mi/h

Average speed,=2v1v2/(v1+v2)

26.5=(2*52.5*v2)/(52.5+v2) on solving we get

V2=17.7mi/h

B)the car returns to the same point so average velocity is 0 mi,/h

C)the average speed will Bethe same as previous 26.5mi/h

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