+138 So I took the derivate of V to get the acceleration, but it ends up with displacement still in the equation I believe, so how do I do this.
The acceleration and velocity are dependent on the displacement ....so the displacement still needs to be in the equation for accel.
+138 okay well how do I solve displacement? The answer for a is a single numerical value, without variables, so I have to be able to solve it somehow.
You are given the displacement of s = 3
put that in your accel equation to calc accel
+138 im talking about question a. the way these problems are supposed to work is a is seperate from b, but b relys on a
This question a) and b) are not dependent....
maximum velocity ...and acceleration at s = 3 are not the same
take derivative of velocity , set = 0 to find s value where velocity is maximum
use this value of 's' in the given velocity equation to calculate the VALUE of the velocity at the 's' you found....
Don't know where ya went....
derivative of the velocity function is the acceleration function....you got that !
derivative of the velocity function also shows the slope of the velocity function at each point
where the slope goes from positive to negative or from negative to positive
the acceleation is too....these are the points where velocity will be a maximum (maybe negative or maybe positive value)
these occur where the acceleration function = 0 (slope = 0 as it goes from pos to neg or neg to pos)
the derivative of the given velocity funtion is given as -10(s^2-4s-4) / (s+4)^2
-10(s^2-4s-4)/(s+4)^2 = 0
-10s^2 +40s+40 = 0 Use Quadratic Formula
s = 2+- 2 sqrt2 these are the points where accel = 0
Plug these values into the orignal Velocity equation to calculate the values......pick the one that is maximum (here I do not know if they want most positive or greatest magnitude)
Use s =3 in the derived ACCELERATION equation to calculate the acceleration at s = 3
The graph above should help to verify your answers......
