A car travels along a straight line at a constant speed of 50.5 mi/h for a dista

Question

A car travels along a straight line at a constant speed of 50.5 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 34.5 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? mi/h (c) What is the average speed for this new trip? mi/h

Explanation / Answer

here,

v1 = 50.5 mi/h

a)

let the constant speed for seccond distance be v2

average speed , v= ( d + d)/( d/50.5 + d/v2)

34.5 = 2 /( 1/50.5 + 1/v2)

v2 = 26.2 mi/h

b)

the net displacement is zero

so the average velocity is also zero

c)

the average speed for this new trip is same as 34.5 mi/h

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