Answer to Question #81172 in Classical Mechanics for Hira

Question

he velocity of a parachutist as a function of time is given by v = vf + (v0 −vf )e
[−t/(2.5s)]
,
where t =0 corresponds to the instant the parachute is opened, v0 = 200 km/h is the veloc-
ity before opening of the parachute, and vf = 18 km/h is the final (terminal) velocity. What
acceleration does the parachutist experience just after opening the parachute?

Answer 1

The acceleration is
a=dv/dt=d/dt (v_f + (v_0 -v_f ) e^[-t/2.5] )=-1/2.5 (v_0 -v_f ) e^[-t/2.5]
The acceleration does the parachutist experience just after opening the parachute:
a(0)=-1/2.5 (v_0 -v_f ) e^[-0/2.5] =-1/2.5 (v_0 -v_f )=-1/2.5 (200 -18 )=-72.8 m/s^2 .