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Homework Help: Trouble deciding on which equation to use in this obscure problem?

  1. Nov 23, 2012 #1
    There are 2 problems, very alike, that I don't know exactly what to do about. I think I may be missing some information that I should know before trying such problems. Any help would be very appreciated.

    1. The problem statement, all variables and given/known data

    The first: A particle P starts at the point O and travels in a straight line. At time t seconds after leaving O the velocity of P is vm s−1, where v = 0.75t^2 − 0.0625t^3.
    (i) the positive value of t for which the acceleration is zero,
    (ii) the distance travelled by P before it changes its direction of motion.

    The second: A particle P moves in a straight line, starting from the point O with velocity 2 ms−1.

    The acceleration of P at time ts after leaving O is 2t^(2/3) m s−2.
    (i) Show that t^(3/4) = 5/6 when the velocity of P is 3 m s−1.
    (ii) Find the distance of P from O when the velocity of P is 3 m s−1.

    2. Relevant equations

    All the suvat equations, probably, should be relevant.

    3. The attempt at a solution

    For the first, we can maybe put a 0 for the a in a = delta v/delta t?
  2. jcsd
  3. Nov 23, 2012 #2
    You can't use the suvat equations here since the acceleration in both problems is not constant.

    You have to use differentiation and integration to solve them, eg in the first problem, you differentiate the expression for v to get an expression for a.
  4. Nov 23, 2012 #3
    This might be a little tough if you haven't seen calculus before but the idea is the following: you know how velocity is rate of change of position? This is expressed as:
    $$ v(t) = \frac{d[x(t)]}{dt} $$
    Notice this is similar to the [itex] v = \frac{\Delta x}{\Delta t} [/itex] equation. That symbol by itself, [itex] v = \frac{d}{dx} [/itex] is an operator (just like, say, √). This means it has no meaning on its own, just when it operates over something. Just like there is a rule for doing the √ operation (taking square root) there is a rule for taking the derivative (that's the name of the operator). It depends on the particular function, but in the case of polynomials (like the ones in your question) the rule is:
    $$ \frac{d[kt^n]}{dt} = nkt^{n-1} $$
    Where k, n are constants. For example, [itex] \frac{d[2t^3]}{dt} = 2\times 3t^{3-1} = 6t^2 [/itex]
    Of course, you can do this multiple times. Taking the derivative of the velocity yields:
    $$ a(t) = \frac{d[v(t)]}{dt} = \frac{d^2[x(t)]}{dt^2}$$
    This just says acceleration is the second derivative of position x(t) and first derivative of velocity v(t). There are many references online for this, one of my favorite is: http://tutorial.math.lamar.edu/

    Edit: I just noticed you also have to do the opposite opeartion of taking derivatives (just like squaring is the opposite operation of √, derivatives have an opposite operation). This is called anti-differentiation (some people call it integration. This is wrong. Indefinite integration is a slightly better term, but it's still a misnomer). As with differentiating (that's a fancy way of saying "taking the derivative") the rule depends on the function, but for polynomials:
    $$ \int kt^n dt = \frac{kt^{n+1}}{n+1} + C $$
    Where k, n, C are constants. It is extremely important you do not forget the C (although for your problems it will not be an issue). Notice the dt is in the numerator now. For example:
    $$ \int 6t^2 dt = \frac{6t^{2+1}}{2+1} = 3t^3 + C$$
    Which is the original function (as we would expect). Now, this means that:
    $$ \int v(t) dt = x(t) + C$$
    $$ \int a(t) dt = v(t) + C$$
    Also, the derivative of a constant is zero (just substitute [itex] kx^0[/itex] in the derivative formula)
    Last edited: Nov 23, 2012
  5. Nov 23, 2012 #4
    Ah, I see. Now I understand. (I haven't covered the calculus part of Mechanics yet.) Thank you both very much indeed!
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