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I know that speed is the derivative of distance. So integrating speed should give you distance. Let's suppose we have a speed which obeys this function:

$$ v(x) = 2^{2^x} $$ So at time 0 the speed is 2 m/s, at time 1 4 m/s, at time 2 16 m/s, and so on.

I would like to calculate the amount of time required to travel a distance $Y$ (let's say 1 km for the sake of the example).

Would I be correct to do this?

$ \int 2^{2^x} dx $ from 0 to t = 1000 m.

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The time derivative of distance is speed. Not just any derivative. – ja72 Sep 30 '15 at 4:44
up vote 2 down vote accepted


First of all, the expression for the velocity that you have there is dimensionally wrong: if $x$ is position, the units of $v$ will turn out to be something$^{metres}$ instead of metres/seconds. So you should stick a constant $C$ in front of your expression so that it takes care of the units, like:

$$ v = C2^{2^x} $$

Velocity = derivative of distance with respect to time, so (in 1D) $ v = \frac{dx}{dt}$;

This is a differential equation, use the method of separation of variables, basically you just consider $dx$ and $dt$ as factors instead of as an operator:

$ v\cdot dt = dx $

$ C2^{2^x} \cdot dt = dx $

$ dt = \frac{dx}{C2^{2^x}} $

and now you integrate:

$\int_0 ^T dt = \int_0 ^Y \frac{dx}{C2^{2^x}} $

so $$ T = \int_0 ^Y \frac{dx}{C2^{2^x}} $$

But that integral does not look nice...

IF YOU MEANT $$ v(t) = 2^{2^t} $$ then I would again suggest sticking a constant in front, so $ v(t) = C2^{2^t} $, and then do the same thing:

$v\cdot dt = dx $

$ C2^{2^t} \cdot dt = dx $

$\int_0 ^Y dx = \int_0 ^T C2^{2^t} \cdot dt$

and you should get $T$ from that, but the integral again does not look nice...

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Based on the description, I believe OP intends for x to be t... – Floris Jul 30 '14 at 3:17
I considered that as well – SuperCiocia Jul 30 '14 at 3:23

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