- #1

Petronius

- 13

- 2

- Homework Statement
- An airboat is a special type of boat that has a large fan mounted on the back instead of a motor that sits in the water. It is used in places where the water is shallow and weedy, so motors that stick down into the water cannot be used.

If the force of the fan pushes the air backwards with 150 N, what is the forward force of the air on the boat? If the mass of the boat is 220 kg, and you ignore friction between the boat and the water, find how far the boat will travel in the first minute.

- Relevant Equations
- Based on unit notes I used:

Faction= -Freaction

F=ma

D= v1∆t + 1/2 at^2

Thank you very much your time!

I first found the force of the air on the boat using the principle of Newton's third law and the fact that no friction is involved.

Faction= -Freaction

150 N backwards = -150 N backwards

150 N backwards = 150 N forward

I then sought to determine the acceleration of the boat forward so that I would have enough information to solve for how far the boat will travel in the first minute (displacement).

F =ma

150 N [forward[ = (220kg)(a [forward])

a = 150 N [forward] / 220kg

a= 0.68m/s^2 |forward|

After this I sought to determine how far the boat would travel in one minute using the kinematics equation D= v1∆t + 1/2 at^2 .

My answer seems far too high and I am a bit perplexed and also wondering if there is an easier equation I could apply

I assumed v1 would be 0 since the the equation seems to state that the boat is starting from rest.

D= v1∆t + 1/2 at^2 .

D = 0 + 1/2(0.68 m/s^2)(60sec)^2

D= (0.34)(3600)

D=1224 metes

Therefore the boat travels a total of 1224 metres in a minute.

I hope my the way I have written the equations is acceptable as they do not seem to directly copy and paste from equation editor on Microsoft word.

Again, thank you for your time and any help/guidance provided.

I first found the force of the air on the boat using the principle of Newton's third law and the fact that no friction is involved.

Faction= -Freaction

150 N backwards = -150 N backwards

150 N backwards = 150 N forward

I then sought to determine the acceleration of the boat forward so that I would have enough information to solve for how far the boat will travel in the first minute (displacement).

F =ma

150 N [forward[ = (220kg)(a [forward])

a = 150 N [forward] / 220kg

a= 0.68m/s^2 |forward|

After this I sought to determine how far the boat would travel in one minute using the kinematics equation D= v1∆t + 1/2 at^2 .

My answer seems far too high and I am a bit perplexed and also wondering if there is an easier equation I could apply

I assumed v1 would be 0 since the the equation seems to state that the boat is starting from rest.

D= v1∆t + 1/2 at^2 .

D = 0 + 1/2(0.68 m/s^2)(60sec)^2

D= (0.34)(3600)

D=1224 metes

Therefore the boat travels a total of 1224 metres in a minute.

I hope my the way I have written the equations is acceptable as they do not seem to directly copy and paste from equation editor on Microsoft word.

Again, thank you for your time and any help/guidance provided.