Register to reply

Acceleration as a function of velocity - HW problem help

by lonsalot
Tags: acceleration, function, velocity
Share this thread:
lonsalot
#1
Jan24-08, 01:31 PM
P: 1
1. The problem statement, all variables and given/known data

The acceleration of a particle is defined by the relation a = -0.05v^2, where a is expresssed in m/s^2 and v in m/s. The particle starts at s=0 m with a velocity of 5 m/s. Determine (a) the velocity v of the particle after it travels 10m, (b) the distance s the particle will travel before its velocity drops to 2 m/s, (c) the distance s the particle will travel before it comes to rest.


2. Relevant equations

a = v (dv/ds)

3. The attempt at a solution



I don't think this is right because when I integrated the right side, it ends up being undefined?
Phys.Org News Partner Science news on Phys.org
Bees able to spot which flowers offer best rewards before landing
Classic Lewis Carroll character inspires new ecological model
When cooperation counts: Researchers find sperm benefit from grouping together in mice
Midy1420
#2
Jan24-08, 03:47 PM
P: 39
not sure if what i'm thinking is correct but its an idea:

a= dv/dt so we know dv/dt = -.05v^2. We can integrate that to get the velocity function.

once we have the velocity function we can integrate again since v(t) = ds/dt. we have initial conditions. its just an idea.
Rainbow Child
#3
Jan24-08, 09:46 PM
P: 365
Quote Quote by lonsalot View Post
...I don't think this is right because when I integrated the right side, it ends up being undefined?
No it isn't!

[tex]\alpha=-0.05\,v^2 \Rightarrow v\,\frac{d\,v}{d\,s}=-0.05\, v^2\Rightarrow \int_{v_0}^v\frac{d\,v}{v}=-\int_o^s 0.05\,d\,s\Rightarrow \ln v\Big|_{v_0}^v=-0.05\,s\Rightarrow v=v_o\,e^{-0.05\,s}[/tex]

jayvastani
#4
Aug20-08, 09:57 AM
P: 1
Acceleration as a function of velocity - HW problem help

Quote Quote by Rainbow Child View Post
No it isn't!

[tex]\alpha=-0.05\,v^2 \Rightarrow v\,\frac{d\,v}{d\,s}=-0.05\, v^2\Rightarrow \int_{v_0}^v\frac{d\,v}{v}=-\int_o^s 0.05\,d\,s\Rightarrow \ln v\Big|_{v_0}^v=-0.05\,s\Rightarrow v=v_o\,e^{-0.05\,s}[/tex]
How can you just cancel "v" on both sides in second step , where it can take a value zero...
Sydney
#5
Nov2-08, 06:47 AM
P: 1
Just to set the record straight on Rainbow child's answer:
No it isn't!

LaTeX Code: \\alpha=-0.05\\,v^2 \\Rightarrow v\\,\\frac{d\\,v}{d\\,s}=-0.05\\, v^2\\Rightarrow \\int_{v_0}^v\\frac{d\\,v}{v}=-\\int_o^s 0.05\\,d\\,s\\Rightarrow \\ln v\\Big|_{v_0}^v=-0.05\\,s\\Rightarrow v=v_o\\,e^{-0.05\\,s}
This slightly wrong, it should be a == d(0.5v2)/ds not d(v2)/ds.
suchara
#6
Sep9-10, 03:04 PM
P: 7
Hi.. just to go one step further, would the distance (with respect to time) equation be :-

[tex]\int{{{v}{}_{0}\over{e^{0.05\,t}}}}\,dt = -{{20\,v{}_{0}}\over{e^{{{t}\over{20}}}\,\log e}}[/tex]
RoyalCat
#7
Sep9-10, 04:21 PM
P: 671
Quote Quote by jayvastani View Post
How can you just cancel "v" on both sides in second step , where it can take a value zero...
Easy, you just say that the step was valid for all non-zero values of v.
[tex]a=v\frac{dv}{dx}=-kv^2[/tex] (In this case, k=0.05)
From this step to the next, you have two options. Either v is always 0, in which case you have a solution for v, and you are not allowed to divide by v. Or v is different from 0, and you are allowed to divide by it.

suchara, the velocity as a function of time is not what you posted. Note that you used the expression for the velocity as a function of distance and replaced the x with t (A mistake).

[tex]v=\frac{v_0}{1+kv_0 t}[/tex]


Register to reply

Related Discussions
Acceleration when velocity is a function of distance Introductory Physics Homework 7
Acceleration/Velocity Problem Introductory Physics Homework 4
Acceleration/velocity problem Introductory Physics Homework 1
Velocity and acceleration problem Introductory Physics Homework 1
Acceleration and Velocity Problem Calculus 2