Optimizing Boat Angle for Hitting a Target Across a River

In summary, the conversation is about trying to find the angle and velocity needed to hit a target on the opposite side of a 400m wide river, while taking into account the river's constant velocity of 0.5 m/s. The person is attempting to use equations involving time and distance, but is unsure if they are doing it correctly and if they need to incorporate the Pythagorean theorem. They are also aware of some errors they have made in their calculations and are seeking advice on how to solve the problem.
  • #1
KurtWagner
43
1
So I have a boat going across a river (y-direction, north) 400m wide.
I am trying to hit a target 75m in the positive x direction on the opposite side of the river.
My boat will travel a velocity v.
The river will provide a constant velocity of +0.5 (east)

I am trying to find the angle a, north of west to point my boat.

hope that all makes sense.

so far I have:

y = vt or t = y/v
t = 400 / v*sin(a)

x = vt
75 = [0.5 - cos(a)] * t

then I plugged in t

75 = 200/[v*sin(a)] - [400*v*cos(a)]/[v*sin(a)]

I can't seem to solve this. Am I doing this the right way?

I was thinking I need another function and I remember having to use
sin(a)^2 + cos(b)^2 = 1
for something like this a while back. And if I do, how do I? Do I just solve the first two equations for sin and cos and then plug the equations into the third, expand them, then solve another for t and then plug it in and simplify?

any help would be greatly appreciated.
 
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  • #2
Hello Kurt, sorry you missed the template. Please use it.

I wonder if your river velocity is 0.5 nautical miles per fortnight or 0.5 m/s, because you missed giving the units. With an error as a consequence:

If you had written your second relevant formula (under 2. in the template) as

75 m = [0.5 m/s - cos(a)] * t

you would have immediately seen that it is nonsense to add a cosine to a velocity (how long is a year plus one ?)

However, this appears to be a typo omission, because under "3. attempt at solution", the v pops up again.

A second error pops up if you temporarily assume the river not to flow. a can then be zero, but instead of t = 400 m / v you get a divergence. You want to interchange cos and sin.

For the rest, you're okay:
vt cos(a) = 400 m plugged into the second eqn gives you
$${75\over 400}v\cos\alpha = 0.5 - v \sin\alpha$$
which is a simple goniometric equation of the form ##a \cos\alpha + b\sin\alpha = c##.
The relevant expression to help you solve that is ##\sin(\alpha+\beta) = ...##
 
  • #3
thanks. sorry for missing the units and for the typo also. I'll give it another crack. :)
 

1. What is a 2d constant motion problem?

A 2d constant motion problem is a physics problem that involves the motion of an object in two dimensions (usually the x and y axes) at a constant velocity. This means that the object is moving at a constant speed and direction without any changes in its motion.

2. How do you solve a 2d constant motion problem?

To solve a 2d constant motion problem, you will need to use the formula: distance = velocity x time. This formula applies to both the x and y directions, so you will need to use it twice to find the distance traveled in each direction. Additionally, you may need to use trigonometry to find the angle of motion and the components of the velocity in the x and y directions.

3. What are some examples of 2d constant motion problems?

Some examples of 2d constant motion problems include a ball rolling off a table, a car moving at a constant speed on a curved road, and a projectile launched at an angle. These problems all involve an object moving at a constant velocity in two dimensions.

4. What are some common mistakes made when solving 2d constant motion problems?

Some common mistakes made when solving 2d constant motion problems include not properly identifying the x and y components of the velocity, not using the correct formula (distance = velocity x time), and not taking into account the effects of gravity on the object's motion.

5. How is a 2d constant motion problem different from a 1d constant motion problem?

A 2d constant motion problem is different from a 1d constant motion problem because it involves motion in two dimensions rather than just one. This means that the object's velocity will have x and y components, and the distance traveled will need to be calculated for both directions. Additionally, trigonometry is often needed to solve 2d constant motion problems, whereas it is not typically needed for 1d constant motion problems.

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