Maximum velocity

  • Thread starter Marioqwe
  • Start date
  • #1
68
4

Homework Statement



A particle starts from rest at the origin and is given an acceleration a=k/(x+4)^2, where a and x are expressed in m/s2 and m, respectively, and k is a constant. Knowing that the velocity of the particle is 4 m/s when x=8m, determine (a) the value of k, (b) the position of the particle when v = 4.5 m/s, (c) the maximum velocity of the particle.

Homework Equations





The Attempt at a Solution



I calculated part (a) and (b) easily. However, I am having some troubles with part (c).
The maximum velocity occurs when the acceleration is equal to zero because there is either a minimum or maximum right? So I set

0 = k/(x+4)^2

which cannot be.
What am I doing wrong?
 

Answers and Replies

  • #2
hunt_mat
Homework Helper
1,741
25
write:

[tex]
v\frac{dv}{dx}=\frac{k}{(x+4)^{2}}
[/tex]

Integrate to get v as a function of x and then write v=dx/dt so find x as a function of t then differentiate again and examine the limit as t tends to infinity.
 
  • #3
68
4
I am getting a very ugly integral. But you said to get v as function of x right?
Then

v = dx/dt ---> dt = (1/v)dx

and integrate the above to get the position x as a function of time? And then take the limit of that function as t goes to infinity? Why would that give me the maximum velocity? Is it because the graph of x(t) has a vertical asymptote?
 
  • #4
hunt_mat
Homework Helper
1,741
25
What do you get for the integral?
 

Related Threads on Maximum velocity

Replies
3
Views
8K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
6
Views
7K
  • Last Post
Replies
1
Views
910
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
944
Top