Hard time with this Vector Velocity Problem

• KevMilan
In summary: Draw a diagram with the paper being the ground. You want to fly north relative to the ground. You might like to draw the air velocity vector first. Then, you should see the direction of the "plane relative to air" vector that is needed to result in a northerly path.In summary, the plane is moving at 200 km/h and the wind pushes it with a velocity of 85 km/h, which results in a resultant velocity of 217 km/h.
KevMilan
Summary: I've posted this in a few forums but still confused on this problem. If the plane is moving at 200km/h and the wind pushes the plane with a velocity of 85km/h, then the resultant velocity would be 217km/h, and using sine inverse, 217.sin(theta=85, I got 23°, which is B, but the answer key shows D.

[Moderator's note: Moved from a technical forum and thus no template.]

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Where did you get the 217km/h?

Did you draw a diagram?

PeroK said:
Where did you get the 217km/h?

Did you draw a diagram?
I used pythagoras theorem, I assumed 200km/h as facing north and 85km/h facing east, and the hypotenuse would be 217km/h.

The hypotenuse should be the 200 km/h in this problem

KevMilan said:
I used pythagoras theorem, I assumed 200km/h as facing north and 85km/h facing east, and the hypotenuse would be 217km/h.

But then the 217km/h is not is a northerly direction (relative to the ground). It is north relative to the air. That's what would happen if you tried to fly north, but didn't take into account the air which is then blowing you eastwards.

Dale said:
The hypotenuse should be the 200 km/h in this problem
A lot of sites have been saying this but I still don't get why the hypotenuse is 200, can you kindly explain this in detail?

KevMilan said:
A lot of sites have been saying this but I still don't get why the hypotenuse is 200, can you kindly explain this in detail?

Draw a diagram with the paper being the ground. You want to fly north relative to the ground. You might like to draw the air velocity vector first. Then, you should see the direction of the "plane relative to air" vector that is needed to result in a northerly path.

With your approach you are defintely going east. Imagine you were trying to walk across a room and I was pushing you sideways. If you try to walk straight, then you'll end up off course. In order to move directly across the room, you would have to walk into me slightly. In this case, and in the case of your problem, the external force is slowing you down, because you are moving into it. That's why the ground speed will be less than 200km/h.

Let ##\vec{e}_x## point from west to east and ##\vec{e}_y## from south to north. Then
$$\vec{v}=v_{\text{wind}} \vec{e}_x + \vec{v}_{\text{plane}}.$$
You want the plane going towards north, i.e.,
$$\vec{e}_x \cdot \vec{v}=0 \; \Rightarrow \; v_{\text{plane} x=-v_{\text{wind}}.$$
Now you only need ##|\vec{v}_{\text{plane}}|=200 \;\text{km}/{\text{h}}## to get ##v_{\text{plane}y}##, and from this result you also get the angle.

KevMilan said:
A lot of sites have been saying this but I still don't get why the hypotenuse is 200, can you kindly explain this in detail?
You have three velocities:

The plane wrt the air (direction unknown, magnitude 200 kph)
The air wrt the ground (direction west, magnitude 85 kph)
The plane wrt the ground (direction north, magnitude unknown)

The last two form a right angle (north and west) so the first must be the hypotenuse.

Dale said:
You have three velocities:

The plane wrt the air (direction unknown, magnitude 200 kph)
The air wrt the ground (direction west, magnitude 85 kph)
The plane wrt the ground (direction north, magnitude unknown)

The last two form a right angle (north and west) so the first must be the hypotenuse.
When you say with respect to air, basically whatever the wind speed is, it'll always be moving at 200km/h in the wind's perspective, right? So the word 'still air' in the question is crucial to finding this solution?

Yes. It is also called “air speed”

Another way to describe what needs to be done: The plane needs a heading such that its East-West component of its 200 km/hr cancels the wind's 85 km/hr.

vanhees71 and PeroK
That's what I've said in #8. All you have to do is to put the numbers in. It also may help a lot to draw the vectors to immediately see, how it works out geometrically (it's simple trigonometry of a right triangle).

1. How do I calculate the vector velocity in this problem?

To calculate the vector velocity, you will need to determine the change in position over a certain time interval. This can be done by finding the difference between the initial and final position vectors and dividing by the time interval.

2. What units should I use for vector velocity?

Vector velocity is typically expressed in meters per second (m/s) or kilometers per hour (km/h). It is important to use consistent units for both the position and time values in order to get an accurate velocity measurement.

3. How do I represent vector velocity graphically?

Vector velocity can be represented graphically as an arrow pointing in the direction of the velocity with a length proportional to the magnitude of the velocity. The direction of the arrow indicates the direction of motion, and the length represents the speed.

4. What is the difference between vector velocity and scalar velocity?

Vector velocity is a measurement that includes both magnitude (speed) and direction, while scalar velocity only includes magnitude. This means that scalar velocity does not account for changes in direction, whereas vector velocity does.

5. How can I use vector velocity to determine acceleration?

Acceleration can be calculated by finding the change in velocity over a certain time interval. This can be done by subtracting the initial velocity vector from the final velocity vector and dividing by the time interval. The resulting vector will represent the acceleration, with magnitude indicating the rate of change and direction indicating the direction of change.

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