At what point does the jumper attain maximum acceleration?

[Sneakatone]

Consider a bungee jumper of mass 65 kg with a 9.00-m cord tied to his ankles. When stretched, this cord may be treated as a spring, of spring constant 150 N/m. Note that the jump-off point is at x=9.00 m, and the origin for the x-coordinate is at the point where the rubber cord becomes taut.
a) At what point does the jumper attain maximum speed?
0=(65kg)(9.81)+150x
x=-4.25

b)what is the maximum speed?
mv^2=kx^2
v=6.3

c)at what point does the jumper attain maximum acceleration?
d) what is the value of this maximum acceleration?

I do not know how to do part c and d.

 

[haruspex]

a) At what point does the jumper attain maximum speed?
0=(65kg)(9.81)+150x
x=-4.25

Yes

b)what is the maximum speed?
mv^2=kx^2
v=6.3

How do you get that equation? The answer is too low.

c)at what point does the jumper attain maximum acceleration?
d) what is the value of this maximum acceleration?

I do not know how to do part c and d.

The force of gravity is constant, so you only have to think about the minimum and maximum of other forces.

 

[Sneakatone]

b) vf^2 - vi^2 = 2ad ( new equation)
v=sqrt(2*-9.81*-4.25)=8.85 m/s^2

 

[haruspex]

b) vf^2 - vi^2 = 2ad ( new equation)
v=sqrt(2*-9.81*-4.25)=8.85 m/s^2

Still too low.
How far has the jumper descended at this point? how much gravitational PE has been lost? How much PE has gone into stretching the cord? What does that leave for the KE?

 

[Sneakatone]

I found v=sqrt(2g(h-x)=16.1

 

[haruspex]

I found v=sqrt(2g(h-x)=16.1

Now you've gone too high. As I indicated in my previous post, some of the lost gravitational PE has gone into stretching the rope.