In what direction must a plane head if it can maintain 350 km/h in still air, wishes to fly 500km in a diirection 040degree and the wind is blowing at 56km/h from 100 degree? howlong would the journey take(to the nearest minutes)? how long would the return journey take, again to the nearest minutes, if the wind is still 56km/h from 100 degree?
I drew the following diagram.
OX is the x-axis, and OY is the y-axis
OB is the desired vector you want to go
so angle(YOB)=40º, it has a magnitude of 500.
Let OA be the vector the plane has to fly, AB is the vector representing the direction and velocity of the wind.
(point A is to the right of B and slightly downwards so it makes a 10º angle with the horizonatal)
Using a bit of basic geometry it should be easy to follow that angle(OBA)=120º
If we let the time taken be t hours, we can label the maginitude of AB as 56t and the magnitude of OA as 350t
Using the cosine law I got the equation
(350t)^2 = 500^2 + (56t)^2 - 2(500)(56t)cos 120º
or
119364t^2 - 28000t - 250000=0
check my arithmetic but by the quadratic formula I got t=1.56 or (a negative t, which we reject)
So the time taken is 1.56 hours
=1.56*60 minutes
=94 minutes to the nearest minute.
Using the Sine Law, Sin(BOA)/(56*1.56) = sin 120/(350*1.56)
sin(BOA)=.1385
angle(BOA)=7.96º
so the direction should be 40º + 7.96º
or appr 48º
Setting up a similar geometrical figure you should be able to do the second part yourself.
To answer the first part of the question, let's break it down step by step:
1. Draw a diagram: Draw a diagram with a reference point O and mark the desired direction as OB, which has a magnitude of 500 km and makes an angle of 40 degrees with the x-axis.
2. Determine the wind vector: The wind is blowing at 56 km/h from 100 degrees. Draw the wind vector AB, which is 56 km/h and makes an angle of 10 degrees with the x-axis.
3. Determine the airplane's vector: Let OA be the vector representing the plane's direction and velocity. We need to find the magnitude and direction of OA.
4. Use the cosine law: Apply the cosine law to the triangle OAB to find OA's magnitude. The cosine law states that c^2 = a^2 + b^2 - 2ab*cos(C), where c is the side opposite angle C. In our case, a = 500 km, b = 56t km (where t is the time taken), and C = 120 degrees (since the angle between OB and OA is 120 degrees). Solving for OA's magnitude gives us (350t)^2 = 500^2 + (56t)^2 - 2(500)(56t)*cos(120).
5. Solve the equation: Simplify the equation and solve for t using the quadratic formula. In this case, the equation becomes 119364t^2 - 28000t - 250000 = 0. Solving for t gives us t = 1.56 hours.
6. Convert time to minutes: To get the journey time to the nearest minute, multiply t by 60 minutes, which gives us 1.56 * 60 = 94 minutes.
Now, to answer the second part of the question, you can follow a similar approach by drawing a diagram and using the same equations, but with the wind still blowing at 56 km/h from 100 degrees. You can solve for the magnitude and direction of OA again and find the return journey time to the nearest minute.