Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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).

I would be correct to do this?

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

EDIT: Yes, I just realized the error!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

No.

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...

share|improve this answer
    
Based on the description, I believe OP intends for x to be t... –  Floris Jul 30 at 3:17
    
I considered that as well –  SuperCiocia Jul 30 at 3:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.