Imagine a ball is moving on the following horizontal line.
. . . . . . . . . . . . . . . . . . . O. . . . . . . . . . . . . . . . . .
Take right as positive. O is the starting point of the ball. Denote the ball by o.
. . . . . . . . . . . . . . . . . . . O. . . . . . . ... . . o . . . . . .
Assume the ball is moving to the right. It has positive displacement since it is on the right of O, and positive velocity since its positive displacement is increasing.
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. . . . . . . . . . . . . . . . . . . O. . . . o . . . . . . . . . . . . .
Now the ball is returning to O. It still has positive displacement since its current position is still on the right of O. However, its velocity is negative since its positive displacement is decreasing and the direction of the velocity vector points left, which is the negative side.
By now you should be able to come up with a scenario where the ball has negative displacement and positive velocity.
You can observe the same phenomenon in daily life. Say, as a stretched spring bounces to its starting position, if we let the returning direction be positive, the string has negative displacement since it is on the negative direction, but has positive velocity. Bungee jump can also used to illustrate the phenomenon.
Answer:
The distance from the North Pole to the equator is 
m.
Explanation:
Circumference of Earth = 40,000 km ......................(1)
Distance from the North Pole to the Equator is = 1/4th of the Circumference of Earth ...................... (2)
Let Distance from the North Pole to the Equator be d ,
the equation formed will be ,
d = 1/4 * Circumference of Earth ........(3)......... ( from equation 1 )
put the value of Circumference of Earth in equation (3),
d = 1/4 * 40,000 km
d = 10,000 km
converting km to m ,
d = 10,000 * 
m
d = 1 * 
m
The distance from the North Pole to the equator is m.
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Answer:
Explanation:
Total weight
My weight+weight of belongings
660+1100=1760N.
a. Work done by the elevator to travel a total height of 15.2m
Using newton law of motion
ΣF = ma
There are only two forces acting upward, the weight and the reaction by the elevator
Also note it is moving at constant velocity then, a=0
N - W=0
Then, N=W
N=1760N
So, workdone is given as
Wordone, =force × distance
Work done=1760×15.2
W=26,752J
W=26.752KJ
b. Work done on me alone is still need to go through the same process but will remove the weight of the belonging
Therefore,
Weight now = 660N
And using the same equation of motion
ΣF = ma
Comstant velocity, a=0
N - W=0
N=W
N=660N
Then, workdone
W=F×d
W=660×15.2
W=10,032J
W=10.032KJ
Explanation:
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