Prove v = u(ktu^3 + 1)^1/3 & x = (1/2ku^2)((ktu^3 + 1)^2/3 - 1)

[SherlockOhms]

Question:
When an object travels through a certain resisting medium the deceleration is proportional to the 4th of the velocity. This, a = -kv^4. Prove v = u(ktu^3 + 1)^1/3 and subsequently x = (1/2ku^2)((ktu^3 + 1)^2/3 - 1).
v at time 0 = u and x at time 0 = 0.

Equations:
Differentiation and integration.

Attempt at solution:
I've proved the first part, for v. I keep getting (1/6ku)((ktu^3 + 1)^2/3 - 1) for c though. So, not too far out. It should just be a simple u substitution following on from the derivation of v. Could anyone walk through the second part? I have no clue where I'm making the mistake.

 

[rock.freak667]

To make things easier to understand (as sometimes you might see 'u' to indicate a variable function for velocity)

ku^3 = A and u=B

So you have

v= dx/dt = B(At+1)^1/3


To solve this you can use another substitution or apply the direct formula for

$$\int (ax+b)^n dx$$

 

[SherlockOhms]

I see where I went wrong. Made a stupid blunder and differentiated the u^3 for some reason. Wasn't thinking. Thanks for that.