Video Transcript
If the position vector of a
particle is expressed as 𝐫 equals two 𝑡 plus three 𝐢 plus five 𝑡 minus two 𝐣,
then the magnitude of the displacement at 𝑡 = 2 seconds equals blank length
units.
Here then we want to fill in this
blank by solving for the magnitude of this particle’s displacement at 𝑡 equals two
seconds. We can begin to do this by
recalling that the displacement vector of an object at a time we’ll call 𝑡 sub f is
equal to that object’s position vector at that same time minus the object’s position
vector at some initial time we’ll call 𝑡 sub i.
In our case, as we want to solve
for displacement magnitude at a time of 𝑡 equals two seconds, we can start by
figuring out the displacement vector at this time. According to our definition, this
displacement vector is equal to the position vector at that same time, two seconds,
minus the position vector at a time 𝑡 equals zero. Knowing this, we can then use our
position vector, which we see is given to us as a function of time, to let us solve
for particle displacement at two seconds.
If we plug in 𝑡 equals two to our
equation for the position vector, we get two times two plus three 𝐢, that’s seven
𝐢, plus five times two minus two 𝐣. That’s equal to eight 𝐣. We can then replace this vector
with the expression we’ve just solved for. Next, we want to solve for the
particle’s position at time 𝑡 equals zero seconds. That’s equal to two times zero plus
three 𝐢, that simplifies to three 𝐢, plus five times zero minus two 𝐣. The vector overall then is three 𝐢
minus two 𝐣. And we substitute this in for 𝐫 of
zero.
We’re now ready to calculate the
displacement at a time 𝑡 equals two seconds. And we get a result of four 𝐢 plus
10𝐣. This, however, is not our final
answer because we want to solve for the magnitude of this displacement. Let’s recall that if we have a
two-dimensional vector, we’ll call it 𝐕, then the magnitude of that vector is equal
to the square root of the sum of the squares of its 𝑥- and 𝑦-components. This means that the magnitude of
our particle’s displacement at 𝑡 equals two seconds equals the square root of four
squared plus 10 squared. Four squared is 16, and 10 squared
is 100. So we’re taking the square root of
116.
116 though is divisible by
four. It’s equal to four times 29. And then, knowing that four is
equal to two squared, we can move it outside the square root sign, which makes our
final answer two times the square root of 29. This is the magnitude of our
particle’s displacement at 𝑡 equals two seconds.
