A car travels along a straight line at a constant speed of 51.0 mi/h for a dista

Question

A car travels along a straight line at a constant speed of 51.0 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 32.5 mi/h.

What is the constant speed with which the car moved during the second distance d?
in mi/h

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? in mi/h

What is the average speed for this new trip? mi/h

Explanation / Answer

case a) Vav=(total displacement)/(total time)

time = time for the first leg(= d/(51.0)) + time for the second leg(= + d/(v)) here v is the speed for the second leg

total displacement = d+ d (in the same direction)

Vav= 32.5 mi/h

32.5 mi/h =2d*/( d/51.0+ d/(v))

1/51 + 1/v = 2/32.5

solving v= 23.8 mi/h

case b) the average velocity is zero in this case, because the displacement is zero, as you have reached the same place from which you started.

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