Motion of a Boat

  • Thread starter Punkyc7
  • Start date
  • #1
420
0
A ferryboat of mass M1 = 2.0 x 10^5 kilograms moves toward a docking bumper of mass M
2 that is attached to a shock absorber. Shown below is a speed v vs. time t graph of the ferryboat from the time it cuts off its engines to the time it first comes to rest after colliding with the bumper. At the instant it hits the bumper, t = 0 and v = 3 meters per second.
a.After colliding inelastically with the bumper, the ferryboat and bumper move together with an initial speed of 2 meters per second. Calculate the mass of the bumper M2

.b.After colliding, the ferryboat and bumper move with a speed given by the expression v = 2e^-4t
. Although the boat never comes precisely to rest, it travels only a finite distance.Calculate that distance.

c.While the ferryboat was being slowed by water resistance before hitting the bumper, its speedwas given by 1/v = 1/3 +βt, whereβ is a constant. Find an expression for the retarding force of the water on the boat as a function of speed.

Homework Equations


1/v = 1/3 +βt
F=ma


The Attempt at a Solution


Im stuck at part C
Im thinking that β would be mu so i get F= 1*10^5((3t-vt/3v)
 

Answers and Replies

  • #2
151
1
So you're given a relationship between velocity and a "retarding force". All forces in Newtonian mechanics can be thought of as accelerations since f=ma. So what you're really trying to find is the acceleration of the boat.

If you are given an expression for velocity, how would you find the acceleration?
 
  • #3
420
0
v/t=a
im not given a time or a distance so how would i solve it?
 
Last edited:
  • #4
151
1
In calculus based physics we would take the derivative of the velocity to find the acceleration. But it looks like you might be in algebra-based physics?

In that case, you have written down two equations for v:

[tex]v=at[/tex]

[tex]\frac{1}{v} = \frac{1}{3} + \beta t[/tex]

And you want an expression in terms of a. So using algebra, you can rearrange, substitute, and solve for a in terms of beta and t.
 
  • #5
420
0
i think its B*t

So solving for a you get (3+Bt^3)/t

So its F= 1*10^5 * (3+Bt^3)/t
 
  • #6
420
0
How would you get a derivative of 1/v = 1/3 +βt
 
  • #7
151
1
i think its B*t

So solving for a you get (3+Bt^3)/t

So its F= 1*10^5 * (3+Bt^3)/t
Hmm...

That's not what I get...Assuming you start from here...

[tex]\frac{1}{at}=\frac{1}{3}+\beta t[/tex]

You can check your answer by substituting what you got for a into the original equation for a (the equation above). You should get the two sides equal to each other if you have the right a, right?
 
  • #8
420
0
a above equals v=at so v/t=a, if you pleg that in the equation you get the same thing as the begining equation 1/v = 1/3 +βt
 
  • #9
151
1
So solving for a you get (3+Bt^3)/t
[tex]\frac{1}{at}=\frac{1}{3}+\beta t[/tex]

To check your answer, let: [tex]a= \frac{3 + \beta t^3}{3}[/tex]

[tex]\frac{1}{t*\frac{3 + \beta t^3}{3}} = \frac{1}{3}+\beta t[/tex]

This simplifes (sort of) to:

[tex]\frac{3}{3t+\beta t^4}=\frac{1}{3}+\beta t[/tex]

Looking at the variable t, we are saying that t to the negative fourth power is roughly proportional to t to the first power. This is only true in very specific circumstances. It is not true generally. In other words, it's like saying that:

[tex]\frac{1}{x^4}=x[/tex]

While you can choose x=1 and it will be true, it's not true everywhere. This is the same kind of problem your solution is giving us.

Start from here:

[tex]\frac{1}{at}=\frac{1}{3}+\beta t[/tex]

And do the algebra again. Try inverting both sides of the equation (to get a in the numerator) and then isolating a by dividing by t...
 
  • #10
420
0
Now im really confused so is it a= 3/t +1/Bt^2
 
  • #11
420
0
how woould you take the derivative of 1/v = 1/3 +βt i know how to do that with regular equation but not with fractions
 
  • #12
151
1
Sorry if I'm confusing you. When I solve it, I get:

[tex]a=\frac{3}{t(1 + 3 \beta t)}[/tex]

To take the derivative of 1/v the easiest thing is to invert it so you can v in the numerator.
 
  • #13
420
0
so if you invert it you get v=3 + Bt^-1 so how would you take the derivative of that im use to seeing x or t and non negative exponets
 
  • #14
420
0
so would it be a=3^-1+ Bt^-2
 

Related Threads on Motion of a Boat

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
0
Views
3K
  • Last Post
2
Replies
29
Views
1K
  • Last Post
Replies
4
Views
7K
  • Last Post
Replies
3
Views
5K
Replies
7
Views
435
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
1
Views
4K
Top