The velocity of a parachutist as a function of time is given by v = vf + (v0 - vf)e-t/2.5, where t = 0 s corresponds to the instant the parachute is opened, v0 = 160 km/h is the velocity before opening of the parachute, and vf = 18 km/h is the final (terminal) velocity.

Question

What acceleration does the parachutist experience just after opening the parachute? (m/s^2)

Answer 1

v = 18 + (160 - 18)e^-(t/2.5)
v = 18 - 142 e^-(t/2.5)
a = dv/dt = (142/2.5) e^-(t/2.5)
when t = 0
a = 142/2.5

Answer 2

whoops,need to change km/hour to meters/second
meters/second = km/hour * 1000 m/km * 1 hr/3600 seconds
= km/hr / 3.6

vo = 160/3.6 = 44.4 m/s
vf = 18/3.6 = 5 /s

then do it