Thread Closed

Motion of a body -calc

 
Share Thread
Sep1-06, 06:43 PM   #1
 

Motion of a body -calc


The motion of a body is given by the equation dV(t)/dt = 0.6 - 3V(t)
where V(t) is the speed (in m/s) at time t (in second). If the body was at rest at t = 0

1) What is the magnitude of the inital acceleration?
2) The speed of the body varies with time as

(A) [tex](1 - e^-^3^t) [/tex]
(B) [tex]2(1 - e^-^3^t)[/tex]
(C) [tex]\frac{2}{3}(1 - e^\frac{-3t}{2})[/tex]
(D) [tex]\frac{2}{3}(1 - e^\frac{-3t}{3})[/tex]

(B) is the correct answer for Q(2) . But how do you arrive at it? And how did they manage to get a 'e' in the answer?

Please help.
PhysOrg.com science news on PhysOrg.com

>> City-life changes blackbird personalities, study shows
>> Origins of 'The Hoff' crab revealed (w/ Video)
>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
Sep2-06, 01:08 AM   #2
 
Recognitions:
Homework Helper Homework Help
Please show what you've tried.

You're dealing with a separable differential equation, do you know how to solve one?
Sep2-06, 02:09 AM   #3
 
what is a separable equation? I know basic calculus. but i have no clue on how to arrive at the answer to this question,
Sep2-06, 03:13 AM   #4
 
Recognitions:
Homework Helper Homework Help

Motion of a body -calc


I googled for "separable differential equation" and found a decent looking text:
http://www.ugrad.math.ubc.ca/coursed...separable.html

Applying the above to your problem:

[tex]\frac{dV(t)}{dt} = 0.6 - 3V(t)[/tex]

[tex]dV(t) = (0.6 - 3V(t))dt[/tex]

[tex]\frac{dV(t)}{0.6 - 3V(t)} = dt[/tex]

[tex]\int_{V_0}^{V}\frac{dV(t)}{0.6 - 3V(t)} = \int_{t_0}^t dt[/tex]

(You could also use indefinite integral, and solve for the C with the information given in the problem ie. "body was at rest at t = 0")


Can you manage the rest?


PS. There's something wrong with the equation or the correct answer. With the given equation you should arrive at:
[tex]0.2(1-e^{-3t})[/tex]

To get the given answer (B), the original equation should be:
[tex]\frac{dV(t)}{dt} = 6 - 3V(t)[/tex]
Sep2-06, 04:50 AM   #5
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus
If you do not know how to solve differential equations, and presumbably aren't expected to here, sSince you are given 4 possible functions, work the other way. Plug each into the equation of motion and see which works. ([itex]\frac{dV}{dt}= 0.6- 3V[/itex] won't work with any of them- as said, it must be 6- 3V.)

As for part A, that's easy. Just evaluate [itex]\frac{dV}{dt}= 0.6- 3V[/itex] at t= 0. (Of course, you are told V(0).)
Sep2-06, 06:24 AM   #6
 
[tex]\int_{V_0}^{V}\frac{dV(t)}{0.6 - 3V(t)} = \int_{t_0}^t dt[/tex]

(You could also use indefinite integral, and solve for the C with the information given in the problem ie. "body was at rest at t = 0")
I am not very sure on how to procede from here.(I have just started learning calculus last week). Anyway should I use
[tex] \int uv =u \int v - \int \frac{(du)}{(dx)}\int v [/tex] rule?
Sep2-06, 08:34 AM   #7
 
Recognitions:
Homework Helper Homework Help
Quote by konichiwa2x
I am not very sure on how to procede from here.(I have just started learning calculus last week).
Here's a formula that should help you:

[tex]\int \frac{dx}{x} = \ln |x| + C[/tex]
Thread Closed

Similar discussions for: Motion of a body -calc
Thread Forum Replies
Resources on three body motion Classical Physics 0
Rigid body motion Advanced Physics Homework 14
rigid body motion Advanced Physics Homework 1
What is rigid body motion? Classical Physics 1
Calc-Motion under gravity Calculus 1