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Thread: C4 Further Differentiation

  1. #1
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    C4 Further Differentiation

    Q: if the velocity v is given by the formula$\displaystyle v = \frac{u}{(1+ks)}$ where u is the initial velocity, s is the distance and k is a constant prove that the accleration varies as $\displaystyle v^{3}$


    my attempt:
    $\displaystyle v \frac{u}{(1+ks)}=\frac{a}{b}$
    $\displaystyle \frac{\mathrm{d}v}{\mathrm{d}s}=\frac{b\frac{\math rm{d}a}{\mathrm{d} s}-a\frac{\mathrm{d} b}{\mathrm{d} s}}{b^2}$
    $\displaystyle a=u; \frac{\mathrm{d} a}{\mathrm{d} s} =0 $
    $\displaystyle b = 1+ks; \frac{\mathrm{d} b}{\mathrm{d} s}= k $


    $\displaystyle \frac{\mathrm{d} v}{\mathrm{d} s}= \frac{((1+ks)(0)-u(k))}{(1+ks)^2}b$


    $\displaystyle \frac{\mathrm{d} v}{\mathrm{d} s}= \frac{-uk}{(1+ks)^2} $




    we know that a = v . dv/ds
    thus; $\displaystyle a = \frac{u}{(1+ks)} \times \frac{-uk}{(1+ks)^2} $


    $\displaystyle a = \frac{-u^{2}k}{(1+ks)^2}$


    note $\displaystyle s = \frac{u-v}{kv} $ sub s into a therefore;
    $\displaystyle a = -\frac{kv^3}{u}$


    thus as v tend to infinity, $\displaystyle v^{3} $ tends to infinity thus acceleration varies as $\displaystyle v^{3} $


    I want to know if I am right?


    Also the bold part is the correct reasoning?
    Last edited by bigmansouf; Feb 28th 2019 at 03:50 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: C4 Further Differentiation

    Quote Originally Posted by bigmansouf View Post
    Q: if the velocity v is given by the formula$\displaystyle v = \frac{u}{(1+ks)}$ where u is the initial velocity, s is the distance and k is a constant prove that the accleration varies as $\displaystyle v^{3}$


    my attempt:
    $\displaystyle v \frac{u}{(1+ks)}=\frac{a}{b}$
    $\displaystyle \frac{\mathrm{d}v}{\mathrm{d}s}=\frac{b\frac{\math rm{d}a}{\mathrm{d} s}-a\frac{\mathrm{d} b}{\mathrm{d} s}}{b^2}$
    $\displaystyle a=u; \frac{\mathrm{d} a}{\mathrm{d} s} =0 $
    $\displaystyle b = 1+ks; \frac{\mathrm{d} b}{\mathrm{d} s}= k $


    $\displaystyle \frac{\mathrm{d} v}{\mathrm{d} s}= \frac{((1+ks)(0)-u(k))}{(1+ks)^2}b$


    $\displaystyle \frac{\mathrm{d} v}{\mathrm{d} s}= \frac{-uk}{(1+ks)^2} $




    we know that a = v . dv/ds
    thus; $\displaystyle a = \frac{u}{(1+ks)} \times \frac{-uk}{(1+ks)^2} $


    $\displaystyle a = \frac{-u^{2}k}{(1+ks)^2}$


    note $\displaystyle s = \frac{u-v}{kv} $ sub s into a therefore;
    $\displaystyle a = -\frac{kv^3}{u}$


    thus as v tend to infinity, $\displaystyle v^{3} $ tends to infinity thus acceleration varies as $\displaystyle v^{3} $


    I want to know if I am right?


    Also the bold part is the correct reasoning?
    You've got your dv/ds correct but I'd advise using "a" in any problem that uses acceleration. It could be very confusing.

    You derived the (correct) relationship between a and v. But you don't need to let anything go to infinity, a is already proportional to v^3 without that. All we need to do to finish the problem is to derive a = (constant)v^3 and we already have that.

    -Dan
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    Junior Member Cervesa's Avatar
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    Re: C4 Further Differentiation

    $\dfrac{d}{dt} \bigg[v = v_0 (1+ks)^{-1} \bigg]$

    $\dfrac{dv}{dt} = -v_0(1+ks)^{-2} \cdot k \cdot \dfrac{ds}{dt}$

    $a = -\dfrac{v_0}{1+ks} \cdot \dfrac{k}{1+ks} \cdot v$

    $a = -\dfrac{v_0}{1+ks} \cdot \dfrac{k}{v_0} \cdot \dfrac{v_0}{1+ks} \cdot v$

    $a = -\dfrac{k}{v_0} \cdot v^3$

    letting the constant $b = -\dfrac{k}{v_0} \implies a = bv^3$

    acceleration varies directly with velocity cubed.
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    Re: C4 Further Differentiation

    well you got the right answer in that

    $a(t) = -\dfrac{k(v(t))^3}{u}$

    thus

    $a(t) \sim (v(t))^3$

    You don't need to bring limits into this
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    Re: C4 Further Differentiation

    Quote Originally Posted by topsquark View Post
    You've got your dv/ds correct but I'd advise using "a" in any problem that uses acceleration. It could be very confusing.

    You derived the (correct) relationship between a and v. But you don't need to let anything go to infinity, a is already proportional to v^3 without that. All we need to do to finish the problem is to derive a = (constant)v^3 and we already have that.

    -Dan
    thank you
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    Re: C4 Further Differentiation

    Quote Originally Posted by romsek View Post
    well you got the right answer in that

    $a(t) = -\dfrac{k(v(t))^3}{u}$

    thus

    $a(t) \sim (v(t))^3$

    You don't need to bring limits into this
    thank you
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    Re: C4 Further Differentiation

    Quote Originally Posted by Cervesa View Post
    $\dfrac{d}{dt} \bigg[v = v_0 (1+ks)^{-1} \bigg]$

    $\dfrac{dv}{dt} = -v_0(1+ks)^{-2} \cdot k \cdot \dfrac{ds}{dt}$

    $a = -\dfrac{v_0}{1+ks} \cdot \dfrac{k}{1+ks} \cdot v$

    $a = -\dfrac{v_0}{1+ks} \cdot \dfrac{k}{v_0} \cdot \dfrac{v_0}{1+ks} \cdot v$

    $a = -\dfrac{k}{v_0} \cdot v^3$

    letting the constant $b = -\dfrac{k}{v_0} \implies a = bv^3$

    acceleration varies directly with velocity cubed.
    thank you for your help
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