(II) Let V₁ (→ above V₁) = ― 6.0&nb...

Question

(II) Let V₁ (→ above V₁) = ― 6.0 î + 8.0 ĵ and V₂ (→ above V₂) = ― 4.5 î ― 5.0 ĵ. Determine the magnitude and direction of

(c) V₁ (→ above V₁) + V₂ (→ above V₂), and

Accepted Answer

Everyone in this problem, we're told to suppose that a velocity vector is represented by vector PKP is equal to 3.9 I hat minus 2.8 J hat meters per second. We're told to then consider another velocity vector Q which is equal to 2.5 I hat plus 4.7 J hat meters per second. And we're asked to find the magnitude and direction of the vector P minus Q. We're given four answer choices each containing a different magnitude or direction for the vector P minus Q that we're looking for. And we're gonna come back to those answer choices as we work through this problem. Now, let's start with the vector P minus Q. OK. We wanna find the magnitude and direction. Let's just start by finding the vector itself. OK? So P minus Q, well, this is just gonna be equal to vector P 3.9. I had minus 2.8 J hat meters per second minus vector Q 2.5. I have plus 4.7 J hat meters per second. I now recall that when we're adding or subtracting vectors like this, we're gonna do it by component. OK. So we're gonna take the I hat components, subtract them, then we're gonna take the J hat components and subtract them. And that's gonna give us the I hat and J hat components respectively of our resulting vector. So we're gonna end up with 3.9 the I A component of the first vector P minus 2.5 the IH A component of this second vector. OK? And that's gonna be our I A component. And then we're gonna add our J hat component and our J hat component is gonna be negative 2.8. OK. So the J hat component of our first vector minus 4.7 the J hat component of our second vector. And this is the J hat component of our resulting vector. And all of this has the units of meters per second. OK. So now we used what we know about vectors to simplify this down. The last step in order to find P minus Q is just to simplify each of these brackets, just a matter of adding and subtracting which we know how to do. So we have 3.9 minus 2.5 that's gonna give us 1.4. So we have 1.4 I have and we have negative 2.8 minus 4.7. So we get minus 7.5 J hot and again, meters per second. OK. So we have P minus Q. Now we need to figure out the magnitude and the direction of this vector. So let's go ahead and label the parts of this vector first. So we're gonna say that the I hat component 1.4 is going to be A and the negative 7.5 is going to be B and we can draw this vector out. So if we draw our axes, OK, this is gonna be 1.4 in the positive X direction, negative 7.5 in the Y direction. So Rector is gonna look something like this. OK. So it's pointing down into quadra four along the X axis. Again, distance of 1.4 and downwards, we have a distance of 7.5. And this makes some angle theta with the X axis. So you can use the formula if you remember the formula for the magnitude of the vector or you can draw it out and see graphically that we can use Pythagorean theorem to find the magnitude of this vector. And the magnitude of this vector is gonna be the length of the hypotenuse of the triangle. We've drawn the magnitude of P minus Q is going to be equal to the square root of A squared. What E squirt. OK. So we're just using Pythagorean theorem here on our A component Nrb component to find the length of that hypo. Let's substitute in our values. We have this square root of 1.4 squared. What negative 7.5 this word. And while we work this out, we get that the magnitude of vector P minus Q is gonna be about 7.62955 m per second. Ok. So that's the magnitude of vector. If we now take a look at our answer choices, let's see if we can narrow anything down. So we're gonna round to two significant digits and our answer is going to be 7.6 m per second for that magnitude. Option A and option C have 2.3 and 2.4 m per second respectively. So those are incorrect. So we can eliminate those. So we're looking at option B or D. Now we need to look at what this angle theta is in order to determine that. So theta again, we can use the formula that we know when we're dealing with vectors or we can look at this graphically and we can see that we can relate the IHAT component and the J hat component through the tangent because they're the opposite and adjacent sides of our triumph. So we're gonna have that tangent of data is going to be equal to the opposite side, which is our B value, the J hap component divided by A. OK. The adjacent side which is our X hack component, we can substitute in these values. So we have 7.5 divided by 1.4. And if we take the inverse tangent of this, we get that theta is going to be equal to about 79.43 degrees. OK. So let's go back up to our answer choices. We have the magnitude and we now have the direction and remember that, that angle theta that we've defined, that's below the positive X axis. OK. So let's go to our answer choices for option B has the correct magnitude and it has a theta is 79 degrees below the positive X axis. Exactly what we found. Option D has 54 degrees above the negative X axis. So that's not correct. So the answer here is option B, we have a magnitude of 7.6 m per second with an angle of 79 degrees below the positive X axis. Thanks everyone for watching. I hope this video helped you in the next one.

Struggling with Physics?

(II) Let V₁ (→ above V₁) = ― 6.0 î + 8.0 ĵ and V₂ (→ above V₂) = ― 4.5 î ― 5.0 ĵ. Determine the magnitude and direction of

(c) V₁ (→ above V₁) + V₂ (→ above V₂), and

Watch next

Unit Vectors practice set