Sorry about the insolence, but I have a little correction to the finish of the video. When the plane will be at east to Bakersfield, it should not change the direction stright to the West becuse the plane will not reach Bakersfield again. the plane will have to flight 8.29° North to West (if the actual speed of the plane remaining 120 m/s) that means velocity of the plane is about 108.75 m/s relative to the ground. And I know this thanks to you :)
The velocities were given in m/sec and the angle must be found using those numbers. The distance was given in miles and thus we used miles for that part of the problem.
why are we allowed to say the velocity of the plane is also the length of the distance from La to Bakersfeild? This does not make sense. Why would this not be a case of similar triangles or some other proof? We can't just state this parameter without probable cause.Please explain :(
You can add and subtract components of vectors. If a plane flied north at 200 km/hour ralative to the air, and a wind blows south at 20 km/hour, the the plane will be flying north at 180 km/hour relative to the ground.
Sorry about the insolence, but I have a little correction to the finish of the video.
When the plane will be at east to Bakersfield, it should not change the direction stright to the West becuse the plane will not reach Bakersfield again.
the plane will have to flight 8.29° North to West (if the actual speed of the plane remaining 120 m/s)
that means velocity of the plane is about 108.75 m/s relative to the ground.
And I know this thanks to you :)
and how we took 120 m/s to 120 miles ?
Is it given in the question that the distance between la and bakersfield is 120 miles or did we find it from the velocity vector of the plane?
The distance from LA to Bakersfield was a "given"
I’m so confused. How do we go from the velocity of 120m/sec to 120 miles. Is it because we’re dealing with relative velocity
The velocities were given in m/sec and the angle must be found using those numbers. The distance was given in miles and thus we used miles for that part of the problem.
@@MichelvanBiezen thx 🙏
why are we allowed to say the velocity of the plane is also the length of the distance from La to Bakersfeild? This does not make sense. Why would this not be a case of similar triangles or some other proof? We can't just state this parameter without probable cause.Please explain :(
When talking about the velocity of a plane, it can be expressed in two ways: 1) relative to the ground 2) relative to the air.
why the x not taking into account the wind?
The vector quantities are added like vectors, and thus all directions are taken into account.
I have a simple solution
Drift distance = time x velocity
= 120 miles /(120-20 sin60)
X (20 cos60) = 11.7 miles
Sir, why we use sin to find the x why not the cos ?
+Naeem Hakimi because cos=adjacent/hypotnuse we need to fin the oposite
please how did you get the angle 30
Glad Noah,
90 degrees - 60 degrees = 30 degrees
I am confused how velocities can be subtracted like 120-17.3
You can add and subtract components of vectors. If a plane flied north at 200 km/hour ralative to the air, and a wind blows south at 20 km/hour, the the plane will be flying north at 180 km/hour relative to the ground.
And the velocity in the y direction where 17.-- was. Found using cos and not sine and treating velocity as distance I mean WOW no comment
The best way to approach these is to represent the velocity as vectors.
we can even use sine and cosine rules for these right?
+Utkarsh Signh I used the sin and cos while solving the problem
Can you check sine rule and cosine rule on net?
he accidentally said wind blowing at 20 miles per second, 72000 mile per hour wind for this problem XD
come on professor get some notes before you do the video, you said 20 miles a second.