Variable acceleration problem

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
Please change your TSR password 23-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
Search Thread
Advanced Search
Sign in to Reply
  1. I am Ace's Avatar
    1 17-07-2012  18:36
    • Peer Of The TSR Realm
    • Posts: 1,447
    Variable acceleration problem
    knowledge required: a= dv/dt=v dv/ds

    Particle moves along straight line with a = 5-2v v=0, t=0
    show that t = \frac {1}{2} ln (\frac{5}{5-2v})s
    and hence express v as a function of t
    show that the velocity of the particle approaches a max value and find this max value
    Last edited by I am Ace; 17-07-2012 at 18:43.
  2. dantheman1261's Avatar
    2 17-07-2012  20:45
    • Full Member
    • Posts: 105
    Re: Variable acceleration problem
    Could you post your attempts thus far?

    Edit: Are you sure the question information is correct? (Think the 's' in the solution shouldn't be there)
    Last edited by dantheman1261; 17-07-2012 at 20:54.
  3. I am Ace's Avatar
    3 17-07-2012  22:37
    • Peer Of The TSR Realm
    • Posts: 1,447
    Re: Variable acceleration problem
    aah, only now I realise that that s will be for seconds! Let me try again and post my solution
  4. I am Ace's Avatar
    4 17-07-2012  22:52
    • Peer Of The TSR Realm
    • Posts: 1,447
    Re: Variable acceleration problem
    Is there a way of saving the latex tags? Instead of having to type them out or copy and paste them?
    Anyway,
    \frac{dv}{dt} =5-2v

    t+c'=\int\frac{1}{5-2v}dv

    t+c'=\frac{-1}{2}\int\frac{1}{u}du

    t+c'=\frac{-1}{2}ln(5-2v)

    c=\frac{-1}{2}ln(5-2(0))=\frac{-1}{2}ln5

    t=\frac{-1}{2}ln(5-2v)+\frac{1}{2}ln5
  5. FireGarden's Avatar
    5 18-07-2012  00:08
    • Exalted Member
    • Posts: 386
    Re: Variable acceleration problem
    That's the same as what they want, but obviously not simplified. It's just laws of logs from there though.
  6. I am Ace's Avatar
    6 18-07-2012  19:45
    • Peer Of The TSR Realm
    • Posts: 1,447
    Re: Variable acceleration problem
    I don't see it,
    I end up with t=\frac{1}{2}ln\frac{5}{5-2v}-\frac{1}{2}
  7. FireGarden's Avatar
    7 18-07-2012  23:31
    • Exalted Member
    • Posts: 386
    Re: Variable acceleration problem
    t = -\dfrac{1}{2} ln(5-2v) + \dfrac{1}{2}ln5 \\ = \dfrac{1}{2}( ln\dfrac{1}{5-2v} + ln5)\\ =\dfrac{1}{2} ln \dfrac{5}{5-2v}
    Last edited by FireGarden; 18-07-2012 at 23:33.
    Post rating:
    1
Sign in to Reply
Search Thread
Advanced Search
Share this discussion:  
Useful resources

Articles:

Study Help rules and posting guidelinesStudy Help Member of the Month nominationsGuide to MathematicsMathematics resourcesMathematics websitesMathematics revision notes in the TSR wikiCambridge Assessment admissions tests (including STEP)How to use LaTeXCareers in Mathematics

Useful Dates:

Quick Link:

Unanswered Maths Threads

Groups associated with this forum:

View associated groups
Article updates
Moderators

We have a brilliant team of more than 60 volunteers looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is moderated by:

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

Shortcuts

Get Started

TSR Group

Using TSR

Info

Connect with TSR

Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.