second derivative problem?
A particle moves in a straight line such that its displacement from a fixed point O, at time t seconds, is given by x metres where
x = 9 ln (1+t) - 4t
Find t when the velocity is numerically equal to the acceleration.
I'm not sure how to do this. I make the first derivative = second derivative but my answer is never right. Its probably an algebra issue or something. any help is appreciated, and please show working 
Comments
Firstly, remember the newtonian Laws of Motion:
If x=dist, v=velocity, a accel.,
v=dx/dt
a=dv/dt
Hence,
v=d(9ln(1+t)-4t)/dt
=[9/(1+t)]-4
a=dv/dt
=-9/(1+t)^2
But v=a
Hence,
[9/(1+t)]-4=-9/(1+t)^2
Let 1/(t+t)=x.
9x-4=-9x^2
9x^2+9x-4=0
9x^2+12x-3x-4=0
(3x-1)(3x+4)=0
x=1/3,-3/4
Putting value of x in x=1/(1+t);
t=2,-7/3
Since t>=0(t cannot be -ve)
Hence t=2s.
It sounds like you are doing it correctly.
velocity is the 1st derivative and acceralation is the second.
after you have that, you would set them equal and solve for "t".
this is what you stated you are doing, so I would check the work from the derivatives.
1st = 9/(t+1) - 4
2nd = -9/(t+1)^2
now set equal and solve
9/(t+1) - 4 = 9/(t+1) - 4
t = -7/4 or t = 2
since this can only be a positive number, it must be t = 2.
x/x-a million: i like product rule greater constructive than quotient rule so im gonna make it x(x-a million)^-a million x(x-a million)^-a million f'(x)= ((x-a million)^-a million)+x(-a million)((x-a million)^-2)(a million) ((x-a million)^-a million)+(-x)((x-a million)^-2) (((x-a million)^-a million)^2)(-x) (-x)(a million/(x-a million))^2 for f''(x) you will possibly only take the spinoff of the respond, which i'm way too lazy to do at 2 something in the morning
Wow sorryi am dumb at a\math