Integrating Velocity When in Unit Vector Notation

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SUMMARY

The discussion centers on the integration of a particle's velocity vector, given by the equation \(\vec{V}(t) = (2t^2 - 4t^3)\hat{i} - (6t + 3)\hat{j} + 6\hat{k}\), to find displacement between \(t=1s\) and \(t=3s\). The integration process is confirmed as correct, yielding a displacement vector of \(-63.3\hat{i} - 30\hat{j} + 12\hat{k}\). The participants agree that each component of the velocity vector can be integrated separately and then summed to obtain the total displacement vector.

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Homework Statement



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

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

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

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



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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.
 
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!
 
Yes, you can.
 

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