Can a boat cross a river along a straight line physics

Set the angle between the bow and the river bank to be , without losing the generality, and then set the shortest distance to cross the river.

On the shore, cos = v1 v2 = 1 2

In **, cos'=v1/v2'=1/3

Therefore, if the captain constantly changes the angle, it is possible to keep the direction of the combined velocity of the ship unchanged.

As shown in the figure, at the beginning, the angle between the bow and the river bank is (sin( +90°-')=2 3), gradually increase , to ** when reached', and then decrease, and the angle is when it reaches the opposite shore.

No, the speed of the boat is too low to be in a straight line when approaching the riverbank.

In order for the boat to cross the river in a straight line, the speed of the boat must be greater than the speed of the water, and then deflect the boat at an angle, which is equivalent to breaking down the speed, one sub-velocity cancels the water velocity, and the other sub-velocity crosses the river.

Of course, the bow of the boat is slightly tilted upstream, so that the sub-velocity of the upward movement just cancels out the speed of the current, and the boat can cross the river in a straight line.

Then you have to see if the captain is willing or not!!

Theoretically, I don't think it will work, but I feel that the refraction of this light is very similar, and the light hits the glass from the air, and the speed slows down, and the angle changes. It's very similar to your question.

I think this kind of problem is boring, but in practice, of course, the velocity component of the vertical and the river bank is as large as possible!

Of course not.

You can decompose the speed of the boat in the direction of the river bank and perpendicular to the river bank, in order to make the boat cross the river vertically, the speed of the water flow should cancel out the speed of the decomposition in the direction of the river bank, but in any case, the speed of decomposition cannot be equal to the speed of the current, so you cannot cross the river vertically.

Suppose v water = v boat = v and the river width is d

When v water = v boat, the shortest displacement is infinitely close to the width of the river, but the closer the displacement is to the width of the river at this time, the longer the time and tends to infinity.

The formula is written as: l>d (not greater than or equal to, but greater than, so infinitely close, not equal to).

And at this time, the time is:

Suppose the ship's offset in the direction of the current is y

vx = d/t

v - vy = y/t

vx^2 + vy^2 = v^2

Solve this equation and eliminate vx and vy to get:

t = (d^2+y^2)/(2v*y)

When y tends to 0, t tends to infinity.

The shortest time is very simple, it is d v, and the actual speed of the boat is 45 degrees in the direction of the river. At this point, the displacement is 2 times the d of the root

The shortest displacement does not exist if V water = V boat, when it is greater than the shortest displacement is the width of the river, and the shortest time is the vertical flow direction of V boat, t = d V boat.

The shortest time is the width of the river v water.

The shortest displacement is infinitely close to the river width.

The shortest time is the width of the river v water.

The shortest displacement is the direction of the vertical water flow.

Hello! Tilting the bow upstream will allow the boat to reach the other side of the river vertically, but for a longer period of time.

The bow is perpendicular to the river bank for the shortest time, but the boat will be biased downstream.

River crossing time = width of the river The minute velocity of the boat perpendicular to the river.

If you don't aim the bow of the boat at the river bank, but deviate from the angle A, the vertical velocity of the boat will be reduced to v*cosa (v is the speed of the boat), and the crossing time will be 1 cosa times the original time. In your question, the vertical water speed does not affect the speed of the boat, and the shortest river crossing time is only related to the speed of the boat, not the speed of the water. The actual displacement is decomposed into vertical displacement and horizontal displacement, and the vertical displacement is the width of the river, which is unchanged.

Then the actual speed of the ship is broken down into two speeds, vertical and horizontal, vertical is the speed of the ship, and horizontal is the speed of the water. No matter how fast the water velocity is, it only affects the displacement in the horizontal direction. The crossing time is removed at a vertical speed.

Hope that helps

a, b, c: because the sub-motion has isochronousness, so when analyzing the river crossing time, only the speed in the direction of the vertical river bank can be analyzed, and when crossing the river, the bow of the boat points vertically to the river bank, that is, the speed direction in the still water points to the river bank, and its size is unchanged, therefore, the boat crossing time is unchanged, option a is correct, and options b and c are wrong

d. When the velocity of the water increases suddenly, the parallelogram rule synthesized by the vector knows the change of the combined velocity of the ship, so the place where the boat reaches the opposite shore changes, and the d option is wrong

Therefore, a

Answer: Water flow speed: 120 feet 10=12 meters;

v: speed of the boat;

v': When the bow of the ship is pointed at an angle to the bank of the Shangyan Trap Kaiyou, the rough velocity of the vertical riverbank;

12^2+v'^2=v^2

10*vTherefore: v=20 m min.

River width: 20*10=200 meters.

This picture?

Analysis: (1) When the boat departs from point A, if the bow of the boat points to point C on the opposite bank of the river, the river crossing time is the shortest.

v1=d/t1=100/(8×60)=

2) Let ab be at an angle to the bank, and it can be seen from the meaning of the title that at this time, it happens to cross to point C on the opposite bank, so the partial velocity of v1 along the bank direction is exactly equal to the river flow velocity v2, so.

v2 v1cos The time to cross the river is at this time.

t=d (v1sin) is obtained by substituting v1=d t1 into the equation.

sinα= t1/t2= 8/10=

So v2=v1cos=

3) If the distance cd of the boat being swept downstream at the second crossing is x, then x=v2t1=v1cos d v1=

First question: Water velocity: 36 6 = 6m s

Speed: [(10 6 20) tg37] 6=10m s

The basic idea of solving this kind of problem is as follows:

Whether you're rowing vertically across the river or diagonally toward the opposite bank, you can always break down the speed of the boat into fractional velocities perpendicular to and along the river. The time taken can be found by the fractional velocity perpendicular to the creek and the width of the creek, and by using this time and the fractional velocity in the direction of the creek, it is possible to find out how far the boat has traveled in the direction of the creek.

It's a simple matter of speed synthesis.

Synthesis and decomposition of velocity.

According to the title, it can be seen that in order to make the boat cross the river in the shortest time, the boat must sail along the water, then the speed of the boat is: the speed of the boat in still water + the flow speed of the river = 3m s + 2m s = 5m s (but here is crossing the river, not sailing along the water or against the current, so don't consider this, the maximum speed is still 3m s).

Answer] t=s v

60m÷3m/s

20s satisfied? I should understand, I don't understand, you can ask again, adopt me).

Of course, it's 20 seconds, and the fastest way to cross the river is to make the boat perpendicular to the river bank The speed component is the largest, that is, the maximum speed is 3

20 It's not against the water or along the water, it doesn't matter what the river is flowing at, it's to the other side.