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

    17
    ReputationRep:


    EDIT TO QUESTION:

    So, I work through part a) using the formula a = d/dx(1/2 v2) which gives me v = 2x1/2(x+3)1/2, which is wrong.

    The answer is v = 2x + 3.

    Not sure how to get there :/
    Offline

    22
    ReputationRep:
    (Original post by Alexion)
    ...
    Do you understand why you're wrong for the first part?

    For the second: velocity is the rate of change of displacement, so write v = 2x +3 \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}t} = 2x + 3 and then it's a simple differential equation.

    For your last question, when the question says that, all it is doing is introducing the variable x as displacement/distance, t as time and v as velocity.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    Do you understand why you're wrong for the first part?
    Not really, I keep getting the same thing as before :/

    For the second: velocity is the rate of change of displacement, so write v = 2x +3 \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}t} = 2x + 3 and then it's a simple differential equation.
    It's easier now I know that the answer to the first bit's simpler, so that's alright now :yep:

    For your last question, when the question says that, all it is doing is introducing the variable x as displacement/distance, t as time and v as velocity.
    Yep, that seems obvious now :rofl:
    Offline

    22
    ReputationRep:
    (Original post by Alexion)
    Not really, I keep getting the same thing as before :/
    You know that acceleration is the rate of change of velocity so you can write a = \frac{\mathrm{d}v}{\mathrm{d}t} then we have that a = 4x +6. But that's in terms of x so we need to use a little trick:

    \displaystyle 

\begin{equation*} a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}x} \, \frac{\mathrm{d}x}{\mathrm{d}t} \end{ equation* }

    since they 'cancel' out. But \frac{\mathrm{d}x}{\mathrm{d}t} is just the rate of change of displacement with time. That's velocity.

    So indeed, we can write \displaystyle a = \frac{\mathrm{d}v}{\mathrm{d}t} = v\frac{\mathrm{d}v}{\mathrm{d}x}.

    We now have a simple DE: \displaystyle a = 4x + 6 \iff v \frac{\mathrm{d}v}{\mathrm{d}x} = 4x + 6 .

    ...looks like your answer in the book is wrong.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    We now have a simple DE: \displaystyle a = 4x + 6 \iff v \frac{\mathrm{d}v}{\mathrm{d}x} = 4x + 6 .

    ...looks like your answer in the book is wrong.
    But how do you get from v(dv/dx) = 4x + 6 to something that's just v = f(x)?
    Offline

    3
    ReputationRep:
    (Original post by Alexion)
    But how do you get from v(dv/dx) = 4x + 6 to something that's just v = f(x)?
    Separating the variables?
    Offline

    22
    ReputationRep:
    (Original post by Alexion)
    But how do you get from v(dv/dx) = 4x + 6 to something that's just v = f(x)?
    Uhm, well \frac{v^2}{2} = 2x^2 + 6x + c , you know that c = \frac{9}{2} so v^2 = 4x^2 + 12x + 9 = (2x + 3)^2, if that's the part you were confused with.

    Otherwise: v \frac{\mathrm{d}v}{\mathrm{d}x} = 4x + 6 \iff \int v \, \mathrm{d}v = \int 4x + 6 \, \mathrm{d}x
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    Uhm, well \frac{v^2}{2} = 2x^2 + 6x + c , you know that c = \frac{9}{2} so v^2 = 4x^2 + 12x + 9 = (2x + 3)^2, if that's the part you were confused with.

    Otherwise: v \frac{\mathrm{d}v}{\mathrm{d}x} = 4x + 6 \iff \int v \, \mathrm{d}v = \int 4x + 6 \, \mathrm{d}x
    OK NOW I SEE WHERE I WENT WRONG

    I subbed in x = 0, v = 0 hence the problems with the resulting equation :facepalm:

    Thanks for the help anyway
    Offline

    22
    ReputationRep:
    (Original post by Alexion)
    ...
    No problem.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What's your favourite Christmas sweets?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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