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?
1
Expert's answer
2018-09-24T12:26:08-0400
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 .
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
Finding a professional expert in "partial differential equations" in the advanced level is difficult.
You can find this expert in "Assignmentexpert.com" with confidence.
Exceptional experts! I appreciate your help. God bless you!
Comments
Leave a comment