A car travels along a straight line at a constant speed of 59.5 mi/h for a dista

Question

A car travels along a straight line at a constant speed of 59.5 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 29.0 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

A) Average velocity = total displacement /total time.   

29 = 2d/(d/59.5 + d/v)

29 = 2*59.5v /(59.5+v)

Solving the equation we get

v = 19.17 mi/h

B) average velocity = 0

because total displacement is zero.

C) average speed = 2d/(d/59.5+d/19.17)

= 29 mi/h

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