The acceleration of the parachute is a function of velocity and equal to A = 8v - 2v^2 + 10. The velocity of the parachute at t = 16 is required.

`A = (dv)/dt = 8v - 2v^2 +10`

At t = 0, as the parachute is dropped its velocity is equal to 0.

`(dv)/dt = 8v - 2v^2 +10`

Look at the graph of the acceleration plotted against velocity:

It is seen that `A = 8v - 2v^2 +10 = (-2v - 2)(v - 5)` . When the value of velocity is v = -1 and v = 5 the acceleration is 0. At t = 0, when the parachute is dropped its velocity is equal to 0 and the acceleration is 10. The value of acceleration remains positive and is increasing till v = 2. Thereafter the acceleration starts to decrease but it is still positive. A positive acceleration increases velocity. The velocity increases further till it reaches v = 5. At v = 5, the acceleration is 0. As a result there is no change in the velocity once it has reached a value of 5.

**The velocity of the parachute is equal to 5 at t = 16.**