Video: Finding the Magnitude of Displacement of a Body at a Given Time given Its Position Expression Relative to Time

If a moving particle has a position vector 𝐫 such that 𝐫(𝑑) = (4𝑑 βˆ’ 3)𝐒 + (6𝑑 + 9)𝐣 in terms of the unit vectors 𝐒 and 𝐣, find the magnitude of the displacement vector of the particle after 2 seconds.

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Video Transcript

If a moving particle has a position vector 𝐫 such that 𝐫 of 𝑑 is equal to four 𝑑 minus three 𝐒 plus six 𝑑 plus nine 𝐣 in terms of the unit vectors 𝐒 and 𝐣, find the magnitude of the displacement vector of the particle after two seconds.

Here we have a vector-valued function. This function describes the position vector of the moving particle. And we’re trying to find the magnitude of the displacement vector. So let’s recall the link between position and displacement. The position vector tells us where the particle is located with respect to the origin at a time 𝑑, whereas the displacement is the change in position of the particle. And so this means that the displacement of the particle after two seconds will be the difference between the position at 𝑑 equals nought seconds and 𝑑 equals two seconds. So that’s 𝐫 of two minus 𝐫 of zero.

We can find 𝐫 of two by substituting 𝑑 equals two into our equation. And we get four times two minus three 𝐒 plus six times two plus nine 𝐣, and that gives us five 𝐒 plus 21𝐣. Then 𝐫 of zero is four times zero minus three 𝐒 plus six times zero plus nine 𝐣, which is negative three 𝐒 plus nine 𝐣. And this means then that the displacement is the difference between these. It’s five 𝐒 plus 21𝐣 minus negative three 𝐒 plus nine 𝐣.

Let’s consider the components individually. We look at the 𝐒-components. We get five 𝐒 minus negative three 𝐒. Well, five minus negative three is eight, so we have eight 𝐒. Similarly, 21 minus nine is 12, so the 𝐣-component is 12. And the displacement is therefore eight 𝐒 plus 12 𝐣. But we’re not quite finished. We were looking to find the magnitude of the displacement, and we remember that a vector has a direction and a magnitude. And so we recall that the Pythagorean theorem can help us find the magnitude of a vector. It’s the square root of π‘Ž squared plus 𝑏 squared.

So in this case, the magnitude of displacement will be the square root of eight squared plus 12 squared, which is the square root of 208. That simplifies to four root 13. Now, we don’t actually have any length units here, so the magnitude of the displacement is four root 13 length units.

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