Calculate Linear Velocity: Aircraft Direction Vector

[Prashant91]

1.A aircraft leaves base and travels west at a speed of 100km/hr for 25 minutes, then turns right and travels north at a speed of 25m/s for 1000 seconds, then turns right and travels east at a speed of 75 km/hr for 10 minutes.

The plane is asked to return to base. Determine the direction that the aircraft must take to return on a direct path and describe this as a position vector.

 

[AlephNumbers]

You have to show an attempt before anyone can help you.

 

[Prashant91]

You have to show an attempt before anyone can help you.

I have converted all the answers to m/s i.e. 100km/hr /3.6 constant = 27.8m/s 25m/s 75km/hr /3.6 constant = 20.8m/s

then converted the time to seconds 25 min = 1500s 1000s 10min=600s

distance = speed *time = 27.8ms*1500=41.7km 25*1000=25km 20.8ms*600secs=12.48 km

then drawing diagram from base position resulted in 41.7km west, 25km north, 12.48km east.

I know i am suppose to use trigonometry but the diagram does not look like a triangle. Please help

 

[AlephNumbers]

You need to find the displacement of the aircraft from the starting point. Are you familiar with unit vectors?

 

[AlephNumbers]

Actually, forget the unit vectors, the cardinal directions work fine. If you travel 10 km west and then 5 km east, how far are you from where you started?

 

[Prashant91]

Actually, forget the unit vectors, the cardinal directions work fine. If you travel 10 km west and then 5 km east, how far are you from where you started?

11.2km away using the pythagorean theroem. I understand i would be able to draw a clear triangle diagram and then find the angle and the distance.

 

[Prashant91]

would i connect the lines between west 41.7 km and east 12.48km to the base position therefore drawing a triangle ?

 

[AlephNumbers]

No, you would be 5 km away from where you started. Here, look at this diagram.

The dotted lines represent the displacement vectors north and west. I do not know if this diagram matches your problem exactly, but it should be something similar.

 

[AlephNumbers]

The arrow on the displacement vector to the west should be going the other way, actually.

 

[Prashant91]

The arrow on the displacement vector to the west should be going the other way, actually.

Okay i think i understand from your diagram. this is because the resultant vector is from the start point to the end point, then you have drawn a triangle. Would my calculations be right in assuming the length of each side. then using pythagorean theorem and tan^-1 to find the angle.

 

[AlephNumbers]

Yes! The actual angle that the aircraft has to turn through would be 90 - θ, though.