Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hi,

    So I'm quite confused about the following Q:

    A particle's motion is governed by the differential equation

    \dfrac{dv}{dt} = -g - k v^2

    where k=\dfrac{\gamma}{m}

    with

    v(t=0) = U > 0

    and \gamma > 0

    v decreases as the particle moves, until v=0 after it has gone a distance D.

    I've tried integrating w.r.t t but nothing seems to be working out.

    The distance is supposed to be int terms of a log by the way.

    Any tips?

    Thanks!
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by golden ratio)
    Any tips?
    One standard technique is to rewrite dv/dt as dv/dx dx/dt = v dv/dx

    And then separate variables and integrate wrt v and x, not t.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by ghostwalker)
    One standard technique is to rewrite dv/dt as dv/dx dx/dt = v dv/dx

    And then separate variables and integrate wrt v and x, not t.
    So

    v \dfrac{dv}{dx} = -g - kv^2

    kv^2 + v \dfrac{dv}{dx}= -g

    \displaystyle \int kv^2 + v \ dv = \displaystyle \int -g \ dx

    and

    \displaystyle \frac{1}{3} kv^3 + \frac{1}{2}v^2 = -gx + C

    But now I'm unsure what to do in order to be able to apply the IC :confused:

    Or is it

    \displaystyle \int \dfrac{v}{-g -kv^2} \ dv = \displaystyle \int 1 \ dx

    ?

    Thanks again!
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by golden ratio)
    Or is it

    \displaystyle \int \dfrac{v}{-g -kv^2} \ dv = \displaystyle \int 1 \ dx
    This. Might be easier to leave the minus's on the right.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by ghostwalker)
    This. Might be easier to leave the minus's on the right.
    I thought so, but then I'm unsure as to how to get t back after I integrate (to apply the IC), because

    x=f(v)
    • Study Helper
    Offline

    16
    ReputationRep:
    Study Helper
    (Original post by golden ratio)
    I thought so, but then I'm unsure as to how to get t back after I integrate (to apply the IC), because

    x=f(v)
    If you find v in terms of x then use v = dx/dt you will end up with another (hopefully simple) DE to solve
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.