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finally i got the question.... now someone solve it
ok i've been complaining bout relative velocity cuz it really just flies past my brain!!!
here's a question... can u solve it with the diagrams n all its just confusing... o n btw someone plz try n make me understand how to solve these sort of qs
In this question, i is a unit vector due east and j is a unit vector due north.
A plane flies from P to Q. The velocity, in still air of the plane is (280i-40j) km/h and there is a constant wind blowing with velocity (50i-70j)km/h. Find
1) the bearing of Q from P
2) the time of flight, to the nearest minute,given that the distance PQ is 273km
and sketching graphs i never can get how u sketch graphs of functions. actually i can do that but wen the question asks you to determine the range/domain of the graph i'm lost!!! i dont want to plot all the points... know any easier ways
thanx....
here's a question... can u solve it with the diagrams n all its just confusing... o n btw someone plz try n make me understand how to solve these sort of qs
In this question, i is a unit vector due east and j is a unit vector due north.
A plane flies from P to Q. The velocity, in still air of the plane is (280i-40j) km/h and there is a constant wind blowing with velocity (50i-70j)km/h. Find
1) the bearing of Q from P
2) the time of flight, to the nearest minute,given that the distance PQ is 273km
and sketching graphs i never can get how u sketch graphs of functions. actually i can do that but wen the question asks you to determine the range/domain of the graph i'm lost!!! i dont want to plot all the points... know any easier ways thanx....
velocity of plane is velocity of plane in still air + velocity of wind
= (280i-40j) + (50i-70j)
=330i-110j (treat i and j like varibles)
use the coefficients as the lenghs of the 2 sides of a right angled triangle and find the angle in between.
angle = tan-1(-110/330) = -18.4 degrees
bearing is turning clockwise so
bearing = 90+18.4= 108.4 degrees
The actual velocity of the plane with respect to the ground is the hypotenuse of that triangle = (110^2+330^2)^0.5 = 347.85 km/h
time is distance over speed = 273/347.85 = 0.78h = 47mins
I hope that helps
= (280i-40j) + (50i-70j)
=330i-110j (treat i and j like varibles)
use the coefficients as the lenghs of the 2 sides of a right angled triangle and find the angle in between.
angle = tan-1(-110/330) = -18.4 degrees
bearing is turning clockwise so
bearing = 90+18.4= 108.4 degrees
The actual velocity of the plane with respect to the ground is the hypotenuse of that triangle = (110^2+330^2)^0.5 = 347.85 km/h
time is distance over speed = 273/347.85 = 0.78h = 47mins
I hope that helps
Oh and for graphs, range is for y and domain is for x. It depends on the equation eg. for y = x^2 domain is all real numbers but range is [0,infinity) because you can't get any x in the domain to make y negative. Also note that the domain usually goes up to real numbers. I have never met a single question asking me to use a domain containing complex numbers before so don't worry about that. After that I guess it is just practice in seeing the graph in your head. Learning how basic transformations (translations, etc) of simple graphs look to get a basic idea of what you are dealing with. I hope it helps.
frixis
Thankyou so much.... btw how are u so clear bout this whole relative velocity could you please tell some basic stuff... you solved it like it was a piece of cake which it is ... not for me though... Thanx ....
OK, all velocity is relative, it's just most are relative to a stationary object ie. the ground (which isn't really stationary since the Earth rotates at thousands of miles per hour but that's another story). If you take velocity relative to a moving object, the relative velocity is the difference between the velocities. So, a car moving at 10mph and another car moving at 10 mph the other way has a relative velocity to the other car of 10- (-10) = 20mph 'cause the other car's going the other way.
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