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I know how , in the physical sense - $$\frac {dv}{dt} = a$$

But, even after thinking a lot, I am not able to see the fault in this - $$\frac {dv}{dt} = \frac {d(st^{-1})}{dt} = \frac {sd(t^{-1})}{dt} = s*(-1)*t^{-2} = \frac {-s}{t^2} = \frac {-v}{t} = -a$$

I know something is terribly wrong here but I'm just not able to figure out what or where. Please keep in mind I'm just a curious 16 year old. Any help would be greatly appreciated.

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    $\begingroup$ v is not s/t. It is ds/dt... assuming s is distance. $\endgroup$ – hft Sep 20 '15 at 18:06
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    $\begingroup$ @hft that should be (expanded into) an answer $\endgroup$ – David Z Sep 20 '15 at 18:09
  • $\begingroup$ Expanded. The problem is that OP only differentiated 1/t not both s and 1/t. $\endgroup$ – hft Sep 20 '15 at 18:14
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I know how , in the physical sense - $$\frac {dv}{dt} = a$$

But, even after thinking a lot, I am not able to see the fault in this - $$\frac {dv}{dt} = \frac {d(st^{-1})}{dt} = \frac {sd(t^{-1})}{dt} = s*(-1)*t^{-2} = \frac {-s}{t^2} = \frac > {-v}{t} = -a$$

I know something is terribly wrong here but I'm just not able to figure out what or where. Please keep in mind I'm just a curious 16 year old. Any help would be greatly appreciated.

Here's an example. Suppose an object is traveling at a constant velocity. Then $$ s=v_0t $$ where $v_0$ is constant in time. I.e., we know that $dv_0/dt=0$ and $ds/dt=v_0$.

So, for this example, you might get confused if you rearrange and write $$ v_0 = \frac{s}{t} $$

and then the RHS looks like it might not be constant... but it has to be, so what gives?

Well, in general: $$ \frac{d}{dt} \left( \frac{s}{t} \right) =-\frac{s}{t^2}+\frac{1}{t}\frac{ds}{dt} $$ because I have to differentiate both terms: the $s$ term; the $\frac{1}{t}$ term. This is just an application of the product rule of differentiation. The above result holds in general.

And, for the example case of constant velocity, plugging in $s=v_0t$, the above equation becomes: $$ -\frac{s}{t^2}+\frac{1}{t}\frac{ds}{dt}=-\frac{v_0t}{t^2}+\frac{1}{t}v_0=0 $$

So, the problem you are having is that you have only differentiated one of the terms, the $1/t$ term, and not the other, the $s$ term.

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  • $\begingroup$ Okay, that clears up a lot. But in your example, you took an object with constant velocity and proved that $dv/dt$ came out to be 0. How would one go about proving that $dv/dt$ actually equals $a$? $\endgroup$ – MayankJain Sep 21 '15 at 11:20
  • $\begingroup$ dv/dt is a, by definition. There is nothing to prove. I've made a minor update to indicate which equations hold generally and which equations hold for the specific example of constant velocity. $\endgroup$ – hft Sep 21 '15 at 19:26

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