Crossing a river ( relative velocity problem )

[Smartguy94]

Homework Statement

A boat is crossing the river with a speed of 6.2 m/s east relative to the water. The river is flowing north at a speed of 2 m/s and is 750 m wide.
a) What is the speed of the boat relative to the Earth?
b) How much time is required to cross the river?
c) How far downriver will the boat land on the other shore?
d) If the pilot of the boat wanted to land directly across the river, how long would the trip take?

Homework Equations

d= rt
others stuff

The Attempt at a Solution

i got a through c quite easily, but I'm stuck with d

a) What is the speed of the boat relative to the Earth? 6.5146 m/s
b) How much time is required to cross the river? 120.9677 s
c) How far downriver will the boat land on the other shore? 241.9355 m
d) If the pilot of the boat wanted to land directly across the river, how long would the trip take?

I'm still having problem with d

this is what I tried

since the horizontal distance is 750m and the vertical distance is 241.9355m
I used pythagorean theorem to find the distance / the hypotenuse and it happened to be 788.0563m

so i used the equation d=rt again and plug in 6.5146 for the r because that's the speed of the boat relative to the earth
788.0563 = 6.5146t
t = 120.9678 s

but its wrong.

 

[Smartguy94]

second try

I try instead of the distance I used in my answer before I use the horizontal distance divided by the velocity of the boat relative to the Earth

d = rt
t = 750 / 6.5146
t = 115.13

but it's still wrong

anyone can help?

 

[Villyer]

In what direction will the boat have to steer in order to move directly across the river?

 

[mizzikz]

I think I have it:
To land directly across the boat would have to travel at an angle that goes against the direction of flow of the river.

If the boat can only move at 6.2 m/s relative to water and its at an angle, then the 6.2m/s is the hypotenuse. To cancel out the motion of the water, it will have to have move 2 m/s in the opposite direction of the flow of water. Then to find the third side:
√(6.2²-2² )= 5.868 m/s relative to the Earth (makes sense that its slower since its gong against the flow)
so then the t=d/v = 750m /5.868 m/s

hope i helped ;)

 

[asteriatic]

I would suggest using the concept of vector addition to solve this problem. The boat's velocity relative to the Earth can be represented as a vector with a magnitude of 6.2 m/s to the east. The river's velocity can be represented as a vector with a magnitude of 2 m/s to the north. To find the boat's velocity relative to the Earth when trying to land directly across the river, we need to find the resultant vector of these two velocities.

Using the Pythagorean theorem, we can find the magnitude of the resultant vector:

√(6.2^2 + 2^2) = √38.24 = 6.18 m/s

To find the direction of the resultant vector, we can use the inverse tangent function:

tan^-1(2/6.2) = 18.4 degrees north of east

Therefore, the boat's velocity relative to the Earth when trying to land directly across the river is 6.18 m/s at an angle of 18.4 degrees north of east.

To find the time required for the trip, we can use the equation d=rt and plug in the horizontal distance of 750 m and the resultant velocity of 6.18 m/s:

750 = 6.18t
t = 121.3 s

So, the trip would take approximately 121.3 seconds.

As for the distance downriver, we can use the equation d=rt and plug in the vertical distance of 241.9355 m and the resultant velocity of 6.18 m/s:

241.9355 = 6.18t
t = 39.2 s

So, the boat will land approximately 39.2 seconds downriver from its starting point.

I hope this helps!