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Problem 6 Easy Difficulty

Using a scale of $1 \mathrm{cm}$ to represent $10 \mathrm{km} / \mathrm{h}$, draw a velocity vector to represent each of the following:
a. a bicyclist heading due north at $40 \mathrm{km} / \mathrm{h}$
b. a car heading in a southwesterly direction at $60 \mathrm{km} / \mathrm{h}$
c. a car travelling in a northeasterly direction at $100 \mathrm{km} / \mathrm{h}$
d. a boy running in a northwesterly direction at $30 \mathrm{km} / \mathrm{h}$
e. a girl running around a circular track travelling at $15 \mathrm{km} / \mathrm{h}$ heading due east



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Video Transcript

{'transcript': "problem. 41. It's labeled Problem 41 in the textbook. What is actually labeled problem 40 but it's actually problem 41. So it is giving us. It is another relative velocity problem, giving us the information that there is a plane heading due south at 600 kilometers per hour and there is a wind coming from the southwest. So if it's coming from the southwest, I mean is going towards the northeast at a speed of 100 kilometers per hour on average, and we are asked to talk like the velocity of the plane relative to the ground. So first, let's draw a diagram. We will say that south and west are negative directions and north of the star positive directions. So strongest diagram. The velocity of the plane relative to the air is 600 kilometers per hour south, so in the negative direction and then the wind is coming from the northeast. So the velocity of the air relative to the ground is 100 kilometers for our north 45 degrees east, and we're looking for the velocity of the plane relative to the ground, and this will give us the direction and This is the magnitude since its velocity. We're looking for both of those. Um so first, let's find the X and Y components of these sides. As you can see for the velocity of the plane relative to the air, it isn't going in the Y direction. So the velocity of the plane relative to the air thanks with the X equals zero kilometers per hour via y equals negative 600 because of the negative Y direction, as we discussed before, 600 kilometers per hour. Now the velocity of the air relative to the ground since it is north 45 degrees east and the sign and co sign of 45 are both equal to each other. That means that the X the X component is going to be equal to the Y component which is equal to 100 coastline 45 which is the same as 100 sign 45 which is approximately 70 point seven kilometers per hour. So these air X and Y components, this is our right component and are exported. So we have all four of those Now we just need to rearrange the Pythagorean theorem to find the magnitude of this which is just the square root of the total X component. So zero plus 70 point seven square plus the total. Why components negative 600 plus 70.7 square and this gives you approximately 540 kilometers per hour has the magnitude. So just is the speed of the plane relative to the ground. But since this is a vector quantity, which is velocity, we need to find the direction as well. So to find data is just the inverse tent of the opposite value over the adjacent. So 70.7 over 529 which he exists an angle of about 7.6 degrees. So the velocity of the plane relative to the ground is 540 kilometers per hour. It is going east. It is going south. Sorry and then turned 7.6 degrees yeast. As you can see here, it's going south and then goes a certain amount, which is data towards the east because the right is the east. Um so that is part A. And this is just the same as 7.6 degrees east of south. Now, that was part a moving on to part B. Part B is much simpler. All it is asking is if how far from the intended position will the plane be after 10 minutes if the pilot doesn't take any corrective action? So if you just let's the wind push the plane, how far from the intended position will be playing beat? So we know that the velocity is just displacement over time. That means displacement is just velocity times time. So since we have the time, that is 10 minutes. But we have all our speeds in kilometers per hour. We should change this two hours, so it's just 1 1/6 hours. So the distance or displacement will be the velocity which we have here. 540 kilometers per hour are sorry. 100 kilometers per hour because it is how far from the intended position will it be? Which is what the wind does. So 100 times, 1/6, and that gives you about 17 kilometers. So that is the answer to part B."}