Ascuba diver makes a slow descent into the depths of the ocean. his vertical position with respect to a boat on the surface changes several times. he makes the first stop 8.0 m from the boat but has a problem with equalizing the pressure, so he ascends 5.0 m and then continues descending for another 12.0 m to the second stop. from there, he ascends 4.0 m and then descends for 14.0 m, ascends again for 8.0 m and descends again for 11.0 m, where he makes a stop, waiting for his buddy. assuming the positive direction up to the surface (+ĵ), express his net vertical displacement vector (in m) in terms of the unit vector ĵ. (express your answer in vector form.)

The total vertical displacement is

Explanation:

Lets say the position of the boat its at the origin of the coordinate system, this is, in vector form:

This simplifies our problem, as the vertical displacement of the diver its:

Now, in this coordinate system, this is the same as:

Be careful. This is only valid in this particular coordinate system.

Ok, we can return to our problem.

Starting at position

He first descends 8.0 m, this is :

,

as the unit vector ĵ points upwards.  Making the substaction:

,

After this, the diver ascends 5.0 m:

,

Then, he continues descending for another 12.0 m:

The, ascends 4.0 m and then descends for 14.0 m, ascends again for 8.0 m and descends again for 11.0 m, all this, will be :

And this is the total vertical displacement.

found in the nucleus

has mass of one amu