This problem looks impossible. If the time depends on the Velocity of the swimmer relative to the river and theta then can't both be interchanged to different values in order to reach across the river at d1 or d2? It almost seems like there are an infinite number of solutions if time to cross the river is not given.
Taking D3 as the hypotenuse and finding it as 250 and then finding sin theta= D2/D3 = Vre/Vsr = 150/250 = 0.6 = 36° 52' 0.6 = Vre/Vsr Vsr = 5/0.6 = 8.33 m/s
but if d2 is equal to (Vre - Vsin(theta)) and d1 is equal to (Vcos(theta))t how can the angle between those two be the same as that between Vcos(theta) and Vsin(theta)?
This problem looks impossible. If the time depends on the Velocity of the swimmer relative to the river and theta then can't both be interchanged to different values in order to reach across the river at d1 or d2? It almost seems like there are an infinite number of solutions if time to cross the river is not given.
Best physics teacher ever ! Thank you so much :))
I think the angle is approximately 16,26 degrees, and Vsr is 5
i sacrificed half my brain space for this so if anyone needs explanation hmu
@@chyres2562 do you still need an explanation?
Taking D3 as the hypotenuse and finding it as 250 and then finding
sin theta= D2/D3 = Vre/Vsr
= 150/250 = 0.6
= 36° 52'
0.6 = Vre/Vsr
Vsr = 5/0.6
= 8.33 m/s
but if d2 is equal to (Vre - Vsin(theta)) and d1 is equal to (Vcos(theta))t how can the angle between those two be the same as that between Vcos(theta) and Vsin(theta)?
I think teta is 45
You might be right. Thanks for chiming in.
Cheers,
Dr. A
t = 30
according to my calculations