Question

A scuba 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.)

Answer 1

The scuba diver's net vertical displacement vector is -38.0 ĵ meters, which means he is 38.0 meters directly below the starting point at the boat.

Step-by-step explanation:

The scuba diver's net vertical displacement can be calculated by adding up the vertical movements. Starting from the boat, the diver descends 8.0 m, ascends 5.0 m, descends 12.0 m, ascends 4.0 m, descends 14.0 m, ascends 8.0 m, and finally descends 11.0 m. Considering the positive direction up to the surface (+ĵ), we can use the vertical displacements with negative values for descents and positive values for ascents.

The net vertical displacement is calculated as follows: (-8.0 m) + 5.0 m + (-12.0 m) + 4.0 m + (-14.0 m) + 8.0 m + (-11.0 m).

When we add these together, we get: -38.0 mĵ. This means that the net vertical displacement of the scuba diver is 38.0 m directly downwards from the boat, which is expressed in vector form as -38.0 ĵ meters.

Answer 2

The total vertical displacement is
`\vec{D}_(diver) = - 28.0 \ m \ \hat{j}`

Step-by-step explanation:

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

`\vec{r}_(boat)=0 * \hat{j}`

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

`\vec{D}=\vec{r}_(diver)-\vec{r}_(boat)`

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

`\vec{D}=\vec{r}_(diver)`

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

Ok, we can return to our problem.

Starting at position

`\vec{r}_(diver) = 0 \ \hat{j}`

He first descends 8.0 m, this is :

`\vec{r}_(diver) = 0 \ \hat{j} - 8.0 \ m \ \hat{j}`,

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

`\vec{r}_(diver) = - 8.0 \ m \ \hat{j}`,

After this, the diver ascends 5.0 m:

`\vec{r}_(diver) = - 8.0 \ m \ \hat{j}  + 5.0 \ m \ \hat{j}`,

`\vec{r}_(diver) = - 3.0 \ m \ \hat{j}`

Then, he continues descending for another 12.0 m:

`\vec{r}_(diver) = - 3.0 \ m \ \hat{j} - 12.0 \ m \ \hat{j}`

`\vec{r}_(diver) = - 15.0 \ m \ \hat{j}`

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 :

`\vec{r}_(diver) = - 15.0 \ m \ \hat{j} + 4.0 \ m \ \hat{j} - 14.0 \ m \ \hat{j} + 8.0 \ m \ \hat{j} - 11.0 \ m \ \hat{j}`

`\vec{r}_(diver) = - 28.0 \ m \ \hat{j}`

And this is the total vertical displacement.