Suppose you drive 1500 m due east in 2 minutes. you then turn due north and drive the same distance in the same time. which of the following is true concerning the average speeds and average velocities for each segment of your trip?
a) the average speeds are different, and the average velocities are different.
b) the average speeds are the same, but the average velocities are different.
c) the average speeds are the same, and the average velocities are the same.
d) the average speeds are different, but the average velocities are the same.

(b) The average speeds are the same, but the average velocities are different

Explanation:

Given,

Distance traveled in the east direction = 1500 mtime take = t = 2 secDistance traveled in the north direction = 1500 mtime taken = t = 2 sec,

Hence the distance traveled in both the direction is the same of 1500 m in the same time of interval t = 2 sec,

We know that the average speed is the ratio of the total distance traveled and the total time is taken. And the speed does not depends upon the direction of the distance traveled. Thus in both cases, average speeds are the same

But in case of the average velocity, average velocity is equal to the ration of the total displacement and the total time is taken, and also it depends upon the direction of the displacement because the velocity is a vector quantity.

In both the magnitude of the case of the velocities are the same but the direction of the velocities is different, due to this the average velocities in both the cases are different.

Therefore the option (b) the average speeds are the same and the average velocities are different, is correct.

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W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j