Isaac Physics Flying Return

https://isaacphysics.org/questions/flying_return?board=cbc6660d-5de6-4207-8783-5be3d57c87d1&stage=a_level

I've been stuck on this question for a while and I can't see what I'm doing wrong.

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Reply 1

mqb2766

Original post by mosaurlodon

https://isaacphysics.org/questions/flying_return?board=cbc6660d-5de6-4207-8783-5be3d57c87d1&stage=a_level

I've been stuck on this question for a while and I can't see what I'm doing wrong.

What did you try?

Reply 2

mosaurlodon

This is the working I've done so far

Reply 3

mqb2766

Original post by mosaurlodon

This is the working I've done so far

As the diagram in hint 3 illustrates, v will not be directed along the from A to B, it must be at an angle to the path to counteract the head/side wind.

Just working it through now.

Reply 4

mqb2766

Original post by mosaurlodon

This is the working I've done so far

Getting isaac to recognise cos as cos, rather than 3 variables ... grrr.

Anyway, think about a velocity isosceles triangle as per hint 3. The base angles are alpha and you can get sin(alpha) in terms of k and sin(theta) using the sin rule. Then project onto the base (direction of motion) to get the velocities as
vcos(alpha) +/-kvcos(theta)
then get the difference in times as youve tried to do.

Reply 5

mosaurlodon

If you have an equation for sin(alpha), how would you find cos(alpha)?

so cos(alpha) would be cos(arcsin(k*sin(theta))) -> is there a rule for this I am unsure how to simplify

Reply 6

mqb2766

Original post by mosaurlodon

If you have an equation for sin(alpha), how would you find cos(alpha)?

so cos(alpha) would be cos(arcsin(k*sin(theta))) -> is there a rule for this I am unsure how to simplify

Its just pythagoras so
sin^2 + cos^2 = 1
so
cos = ...

Just think of a unit (hypotenuse) right triangle with base cos and height sin and everything falls out of that.

Reply 7

mosaurlodon

so the time difference should be x/v - x/2vkcos(theta) right?

Reply 8

mqb2766

Original post by mosaurlodon

so the time difference should be x/v - x/2vkcos(theta) right?

No, youd have (with appropriate substitution)
v1,v2 = vcos(alpha) +/- kvcos(theta)
so
x/v1 - x/v2
and simplify it. If thats what youve tried to do, post your working.

(edited 2 months ago)

Reply 9

mosaurlodon

sorry to have this take so long, but are my velocities correct?

so Vab = vcos(alpha) + kvcos(theta) = v(2kcos(theta))
and Vba = vcos(alpha) - kvcos(theta) = v

Reply 10

mosaurlodon

-> from the equation sin(alpha) = k*sin(theta)
so cos(alpha) = sqrt(1-k^2sin^2(theta)) = kcos(theta)

Reply 11

mqb2766

Original post by mosaurlodon

sorry to have this take so long, but are my velocities correct?

so Vab = vcos(alpha) + kvcos(theta) = v(2kcos(theta))
and Vba = vcos(alpha) - kvcos(theta) = v

No, alpha and theta are not the same angle. The headwind is at an angle theta, the driving velocity is at an angle alpha, both relative to the resultant direction of motion.

You "know" cos(theata), but have to get an expression for cos(alpha) in terms of sin(theta) as per the previous posts.

Reply 12

mqb2766

Original post by mosaurlodon

-> from the equation sin(alpha) = k*sin(theta)
so cos(alpha) = sqrt(1-k^2sin^2(theta)) = kcos(theta)

cos(alpha) = sqrt(1-k^2sin^2(theta))
is correct. Its not equal to kcos(theta).

(edited 2 months ago)

Reply 13

mosaurlodon

Oh my gosh, I can't believe it my answer was right

Thank you so much!!

Reply 14

mosaurlodon

my mistake was k^2sin^2(theta)+k^2cos^2(theta)=1 rather than k^2sin^2(theta)+k^2cos^2(theta)=k^2

Reply 15

mosaurlodon

ok I was working on part b of this question, and my monkey brain decided to randomly replace theta with phi, and it somehow turned out to be the right answer - can someone explain this to me?

Reply 16

mqb2766

Original post by mosaurlodon

ok I was working on part b of this question, and my monkey brain decided to randomly replace theta with phi, and it somehow turned out to be the right answer - can someone explain this to me?

Can you post what you did?

Reply 17

mosaurlodon

I also am confused on which way the wind is blowing, in the q, it says from north to south, but in the hint 2, it says from west to east.

Reply 18

mqb2766

Original post by mosaurlodon

I also am confused on which way the wind is blowing, in the q, it says from north to south, but in the hint 2, it says from west to east.

Ok, just wanted to make sure. The hint 2 part is wrong, and really if you rotated the diagram (hint 3) round from part a) so that the wind was blowing northerly and phi is the bearing (relative to north), then its just theta from part a).

Most of the question part info is irrelevant, you simply need the direction of the headwind relative to your direction of motion (as you should expect).