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

Video Transcript

A moving particle has a position vector 𝐫 given by the relation 𝐫 of 𝑡 equals six 𝑡 minus four 𝐢 plus nine 𝑡 plus four 𝐣, where 𝐢 and 𝐣 are the unit vectors. Find the magnitude of the particle’s displacement during the interval two to six seconds.

We need to be really careful when answering this question. We’ve been given information about the position of the particle 𝐫 at time 𝑡. And we’re looking to calculate the particle’s displacement. So what is the difference between position and displacement? The position tells us the exact location with respect to a fixed point, here, the origin, whereas the displacement tells us the distance traveled in a particular direction from that origin.

And so we’re going to begin by letting 𝑡 be equal to two and then 𝑡 be equal to six and calculate the exact position of the particle at these times. Once we’ve done that, we’ll be able to calculate the displacement. 𝐫 of two is six times two minus four 𝐢 plus nine times two plus four 𝐣. That gives us a position vector of eight 𝐢 plus 22𝐣 when 𝑡 is equal to two. Next, we let 𝑡 be equal to six, and we get six times six minus four 𝐢 plus nine times six plus four 𝐣, giving us 32𝐢 plus 58𝐣.

We want to calculate the displacement vector over this interval. And that will be given by its change in position over this time, so 𝐫 of six minus 𝐫 of two. That’s 32𝐢 plus 58𝐣 minus eight 𝐢 plus 22𝐣. And if we distribute this negative over the entire parentheses, we get 32𝐢 plus 58𝐣 minus eight 𝐢 minus 22𝐣. We can now subtract the individual components. 32𝐢 minus eight 𝐢 is 24𝐢, and 58𝐣 minus 22𝐣 is 36𝐣. So the change in position, and therefore the displacement vector, is 24𝐢 plus 36𝐣.

We’re not quite finished though. We want to calculate the magnitude of the displacement. And so if we consider the displacement vector as a single straight line, we know we can represent this as a right-angled triangle as shown. The magnitude of the displacement is simply the length of this line. And since we’re working with a right-angled triangle, we can use the Pythagorean theorem. This tells us that the square of the longest side in our triangle is equal to the sum of the squares of the two shorter sides. In other words, 𝑥 squared is equal to 24 squared plus 36 squared. 24 squared plus 36 squared is 1872. So 𝑥 will be the square root of this, which is 12 root 13. The magnitude of the particle’s displacement is, therefore, 12 root 13.