Hope this helped.-Im only gonna do 1. Also, what is that I in 1? x(t) = ln(I + t^2) ? Whats that I?......
109 = c or
109 = 5c^3 - 6
(109 + 6)/5 = c^3
23 = c^3
23^(1/3) = c
7.667 = c.
4.
V = w * h * L
36 = 4 * h * L
9 = h * L
9/h = L
Go on from here. I'm going to sleep. Night.
Hope this helped.
I'm only gonna do 1. Also, what is that I in 1? x(t) = ln("I" + t^2) ? What's that I? Im' gonna assume its a 1.... Unless you wanna edit and maybe i'll come back.
So just take the second derivative of that position function and you will set that equal to 0 to obtain a value for t where acceleration is 0. Then plug in that t value for the first derivative of the position function, which is velocity.
So the first derivative is : 2t / (1+t^2)
Second derivative: ( 2(1+t^2) - 4t^2 ) / (1+t^2)^2
so you gotta set the numerator of the 2nd derivative equal to 0, which should give you a t value where the acceleration is 0. so:
-2t^2 + 2 = 0 => -2(t^2 - 1) = 0 => -2 (t+1)(t-1) = 0 so at time t=1 acceleration will be 0. You don't take t=-1 because you can't have negative time.
So take t=1 and plug it into the first derivative of the position function which is above. 2(1) / (1 + (1)^2). Which is just 1....
