• Support PF! Buy your school textbooks, materials and every day products Here!

Short calculus task

  • Thread starter vs55
  • Start date
  • #1
21
0

Homework Statement


A particle initially moving at 5m/s has an acceleration of a=-.5v.
Find v(t)


Homework Equations





The Attempt at a Solution


i know the anti derivative of acceleration is -(v^2)/4
but i have no idea how to put velocity in terms of time when acceleration in terms of v is given
 

Answers and Replies

  • #2
1,341
3
It's a separable differential equation:

[tex]a = \frac{dv}{dt} = -\frac{1}{2}v[/tex]

Does that help?
 
  • #3
21
0
not really...i still don't fully understand how to put v as a function of time
 
  • #4
When solving a separable differential equation, you treat the derivative (dv/dt) as a fraction. Get all the v's and dv's on one side and all the t's and dt's on the other side:
dv/dt=-1/2 v
1/v dv=-1/2 dt

Then integrate both sides and solve for v.
 
  • #5
21
0
the answer given in the notebook is "v=5 exp(-.5t)"...i don't understand :s
when i integrate i get
ln|v| = -1/2t
solving for v = e^(-.5t)
where is this 5 coming from??
 
Last edited:
  • #6
You forgot the constant in your integration. What you have is an initial condition problem: v(0)=5. So when you integrate both sides, you get
ln|v| = -1/2 t + c
v=e^(-1/2 t+c)

It makes things neater to get that constant out of the exponent, using the laws of exponents:
v=e^c*e^(-1/2 t)

Now, since c is a constant, e^c is a constant, so we can just cal lthat a new constant to make it neater:
Let k=e^c
v=k*e^(-1/2 t)

With your initial condition v(0)=5:
5=k*e^(-1/2*0)
5=k*1
k=5

So v=5e^(-1/2 t)
 
  • #7
21
0
ahh i c thx alot!
 

Related Threads for: Short calculus task

  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
461
  • Last Post
Replies
1
Views
539
  • Last Post
Replies
5
Views
494
  • Last Post
Replies
4
Views
2K
Top