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Question :

if $s^2$ is a quadratic function of $t$, prove that the acceleration varies inversely as $s^3$.


So we have that

$$ s^2 = at^2 + bt + c $$

And want to show that

$$ a = c s^{-3} $$


If $s^2 = at^2 + bt + c$ then $v = 2s = 2at + b$

Which also means that $v = 2 \sqrt{at^2 + bt + c}$

Then acceleration is

$$ \frac{dv}{dt} = \left( at^2 + bt + c \right)^{-1/2}(2at + b) $$

Or

$$ a = \frac{ (2at + b) }{ \left( at^2 + bt + c \right)^{1/2} } $$

Subbing in $s$ gives

$$ a = \frac{ (2at + b) }{ \left( s^2 \right)^{1/2} } $$

Or

$$ a = \frac{ (2at + b) }{ s } $$

But $v = (2at + b) = 2s$ , so

$$a =\frac{v}{s} =\frac{2s}{s}=2 $$

So $a = 2$?


which is clearly wrong, but I'm not sure where or why.


ans:

We have

$$ s^2 = \alpha t^2 + \beta t + \gamma $$

Differentiating this we get

$$ 2s \frac{ds}{dt} = 2 \alpha t + \beta $$

Using that $v = \frac{ds}{dt}$ this means

$$ 2s v = 2 \alpha t + \beta $$

Dividing through by $2s$ gives an expression for $v$ as

$$ v = \frac{2\alpha t + \beta}{2s} $$

Using the quotient rule to differentiate velocity and find acceleration gives

$$ a = \frac{ (2 \alpha)(2 s) - 2v (2\alpha t + \beta) }{ 4 s^2 } $$

If we sub in $v = \frac{2\alpha t + \beta}{2s}$ to the above we get

$$ a = \frac{ (2 \alpha)(2 s) - 2 \left( \frac{2\alpha t + \beta}{2s} \right) (2\alpha t + \beta) }{ 4 s^2 } $$

Which gives

$$ a = \frac{ (2 \alpha)(2 s) - \left( \frac{1}{s} \right) (2\alpha t + \beta)^2 }{ 4 s^2 } $$

Combining terms gives

$$ a = \frac{ 4 \alpha s - \left( \frac{ (2\alpha t + \beta)^2 }{s} \right) }{ 4 s^2 } $$

$$ a = \frac{ \frac{ 4 \alpha s^2 - (2\alpha t + \beta)^2 } {s} }{ 4 s^2 } $$

Which reduces to

$$ a = \frac{ 4 \alpha s^2 - (2\alpha t + \beta )^2 } { 4 s^3 } $$

From here we can sub in $s = \sqrt{\alpha t^2 + \beta t + \gamma}$ and expanding the terms gives

$$ a = \frac{ 4\alpha^2 t^2 + 4 \alpha \beta t + 4 \alpha \gamma - 4\alpha^2 t^2 - 4\alpha \beta t - \beta^2 } { 4 s^3 } $$

Simplifying to:

$$ a = \frac{ 4 \alpha \gamma - \beta^2 } { 4 s^3 } $$

Which proves that the acceleration varies inversely as $s^3$ as needed.

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    $\begingroup$ $v=\frac{ds}{dt}$, not what you wrote. $\endgroup$ – Damian Sowinski Apr 28 '17 at 13:35
  • $\begingroup$ @DamianSowinski ah - I have got acceleration where I should have velocity, it seems? $\endgroup$ – baxx Apr 28 '17 at 13:38
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So we have that

$$s^2 = at^2 + bt + c$$

For fun, try a more general approach. Let $$s = \sqrt{f(t)} \rightarrow \dot s = \frac{\dot f(t)}{2\sqrt{f(t)}} \rightarrow \ddot s = \frac{\ddot f(t)}{2\sqrt{f(t)}}-\frac{\dot f(t)^2}{4f(t)^{3/2}} = \frac{f(t)\ddot f(t)-\frac{1}{2}\dot f(t)^2}{2f(t)^{3/2}}$$

Recognize that $f(t)^{3/2} = s^3$ and then write

$$\ddot s = \frac{f(t)\ddot f(t) - \frac{1}{2}\dot f(t)^2}{2s^3}$$

So we seek a function $f(t)$ such that the numerator is a constant

$$f(t)\ddot f(t) - \frac{1}{2}\dot f(t)^2 = K$$

Certainly, $f(t) = kt$ solves this differential equation and then check that

$$s = \sqrt{kt} \rightarrow \dot s = \frac{k}{2\sqrt{kt}}\rightarrow \ddot s = -\frac{k^2}{4(kt)^{3/2}} = -\frac{k^2}{4s^3}$$

as desired. Let's see if a quadratic works

$$f(t) = at^2 + bt + c \rightarrow \dot f = 2at + b \rightarrow \ddot f = 2a$$

Plug these in to the differential equation

$$(at^2 + bt + c)(2a) - \frac{1}{2}(2at + b)^2 = 2ac - \frac{b^2}{2} = K$$

and see that a quadratic is also a solution. Let's check

$$s = \sqrt{at^2 + bt + c} \rightarrow \dot s = \frac{2at + b}{2\sqrt{at^2 + bt + c}} \rightarrow \ddot s = \frac{4ac -b^2}{4\sqrt{at^2 + bt + c}} = \frac{4ac -b^2}{4s^3}$$

as desired.

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After you differentiate displacement .. You will get 2sv= 2at+b Here v represents the velocity. Thus, v= 2at+b/2s Differentiating velocity wrt time we will get acceleration. We can use quotient rule for differentiating RHS . Then we simple solve and simplify it giving us. a = 4ac-b^2/2s^3 Hence proved

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  • $\begingroup$ They've given it as $s = $ displacement, so that I would have $velocity = (1/2)(at^2 + bt + c)^{-1/2}(2at + b) $ wouldn't I? $\endgroup$ – baxx Apr 28 '17 at 13:53
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    $\begingroup$ Don't put the value of displacement just yet.. Leave it as s then differentiation becomes easier $\endgroup$ – Lakshya Gupta Apr 28 '17 at 13:55
  • $\begingroup$ Ok thanks - but then I have that $v = \frac{2at + b}{2s}$ giving $a = \frac{2a(2s) - (2at + b)(2)}{4s^2}$, so I'm not sure what I've missed there $\endgroup$ – baxx Apr 28 '17 at 14:00
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    $\begingroup$ You did not take the LCM of the numerator properly. It would be 4s^2a $\endgroup$ – Lakshya Gupta Apr 28 '17 at 14:01
  • $\begingroup$ I don't follow, you're using LCM = lowest common multiple? And you're saying that it would be 4 * s^{2a} , ? It's hard to understand what you're mean as you're using no parenthesis in your expressions for operation orders $\endgroup$ – baxx Apr 28 '17 at 14:04
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You have $s(t)=\sqrt{at^2+bt+c}$. Now calculate $d^2s/dt^2$ and you will get the result after substituting $\sqrt{at^2+bt+c}=s$ back.

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