Can the tanker avoid hitting the reef with its engines back on?

[ledhead86]

An oil tanker's engines have broken down, and the wind is blowing the tanker straight toward a reef at a constant speed of 1.5 m/s. When the tanker is 500 m from the reef, the wind dies down just as the engineer gets the engines going again. The rudder is stuck, so the only choice is to try to accelerate straight backward away from the reef. The mass of the tanker and cargo is 3.6*10^7 kg, and the engines produce a net horizontal force of 8*10^4 N on the tanker.

Will the ship hit the reef? You can ignore the retarding force of the water on the tanker's hull.

Here is a diagram of the problem
http://image61.webshots.com/161/8/51/92/474485192tnwxrf_ph.jpg"

 

[ledhead86]

I know the acceleration would be -.0033333333 by the formula: a=Σ F/m
= 8*10^4 / 3.6*10^7.
Where do I go from here?

 

[ledhead86]

I reposted the question with a picture.

 

[Skippy]

At what time does the tanker come to a stop?

Use that time to see how far it went.

 

[ledhead86]

I didn't know it came to a stop?

 

[Skippy]

No, assume the reef isn't there when you calculate how long it takes the tanker to stop.

Then see how far it goes using the time you calculated. If D > 500 then it hits...

 

[ledhead86]

how do i do that?

 

[Skippy]

The equation is V = a*t. You said a= -.0033. Also, you know you're traveling at 1.5m/s. So you want to know when your acceleration will get you to -1.5 m/s (so it will cancel out with the foward motion and you will stop).

 

[rmedrano89]

the answer is YES, it will hit the reef

 

[rocksolid99]

I found this thread while googling this same question, so I just wanted to add a few notes I found:

I know the acceleration would be -.0033333333 by the formula: a=Σ F/m
= 8*10^4 / 3.6*10^7.
Where do I go from here?

The acceleration is -2.22e-3, not sure how you got -3.33...

The equation is V = a*t. You said a= -.0033. Also, you know you're traveling at 1.5m/s. So you want to know when your acceleration will get you to -1.5 m/s (so it will cancel out with the foward motion and you will stop).

0 should be your final velocity, not -1.5. A negative acceleration, not velocity, will result in the boat eventually stopping. I got 506.7m for the total distance to stop, and my version of the problem asks what speed the boat hits at too: .17m/s for the velocity at which the boat hits the reef, which satisfies the hull's limit of .2m/s to maintain integrity.

-bf