# Thread: C4 Further Differentiation

1. ## 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?

2. ## Re: C4 Further Differentiation Originally Posted by bigmansouf 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

3. ## 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.

4. ## 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

5. ## Re: C4 Further Differentiation Originally Posted by topsquark 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

6. ## Re: C4 Further Differentiation Originally Posted by romsek 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

7. ## Re: C4 Further Differentiation Originally Posted by Cervesa $\dfrac{d}{dt} \bigg[v = v_0 (1+ks)^{-1} \bigg]$

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

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

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

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

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

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