Integrating Velocity When in Unit Vector Notation

[ThomasMagnus]

Homework Statement

Say for example, a particles velocity was given by the following equation:

[itex]\vec{V}[/itex](t) = (2t²-4t³)[itex]\hat{i}[/itex] - (6t +3)[itex]\hat{j}[/itex] + 6[itex]\hat{k}[/itex]

If I wanted to find the displacement of the particle between t=1s and t=3s, could I just integrate like this?

[itex]\int \vec{V}[/itex]= (2t³/3 - t^4)[itex]\hat{i}[/itex] - (3t² +3t)[itex]\hat{j}[/itex] + 6t [itex]\hat{k}[/itex] evaluated between 1.00 and 3.00

= (-63i)-36j + 18k)-(2/3-1)i+(6j)-6k= -63.3i - 30j + 12k.

Is this the correct way to do this?

Homework Statement

N/A

Homework Equations

N/A

 

[Maybe_Memorie]

Yep, that's correct.

As for why it's correct, suppose the particle's velocity was just 6i, so the distance is only changing in the i direction so you only integrate in that direction. Then if it's velocity was 6i + 3j, the total displacement is the same as moving the i component, then traveling in the j component separately.

The total displacement is just the vector sum, hence why your integration is correct.

 

[ThomasMagnus]

So can you just treat each unit vector separately and integrate and evaluate each individually, then combine them all to find the displacement vector?

Thanks!

 

[Maybe_Memorie]

Yes, you can.

 

[HalfLostAndSearching]

The integration of velocity in unit vector notation is a valid method for finding the displacement of a particle between two given times. Your approach is correct, as you have integrated the velocity components in each direction and evaluated them at the given times. However, it is important to note that the displacement vector is the integral of the velocity vector, so the units of your final answer should be in meters (m) or whatever unit your position vector is in. Additionally, it may be helpful to include units in your calculations to ensure accuracy. Overall, your approach is correct and you have successfully found the displacement of the particle between t=1s and t=3s.