Question

3. Elementary Mechanics (a) Write an initial value problem to model the velocity of a 2 pound object with initial velocity 2 feet per second upwards. Assume that air resistance is proportional to speed, with k = 0.011bs/ft. Make up the positive direction. (b) Write an initial value problem to model the velocity of a 2 pound object with initial velocity 2 feet per second upwards.. Assume that air resistance is proportional to speed, with k = 0.011bs/ft. Make down the positive direction. (C) Write an initial value problem to model the velocity of a 2 kilogram object that is released from rest. Assume that air resistance is proportional to speed, with k = 0.01N s/m. Choose your own coordinate system but make sure to clearly define with direction is the positive direction.

3. Elementary Mechanics

(a) Write an initial value problem to model the velocity of a 2 pound object with initial velocity 2 feet per second upwards. Assume that air resistance is proportional to speed, with k = 0.01lb · s/ft. Make up the positive direction.

(b) Write an initial value problem to model the velocity of a 2 pound object with initial velocity 2 feet per second upwards.. Assume that air resistance is proportional to speed, with k = 0.01lb · s/ft. Make down the positive direction.

(c) Write an initial value problem to model the velocity of a 2 kilogram object that is released from rest. Assume that air resistance is proportional to speed, with k = 0.01N · s/m. Choose your own coordinate system but make sure to clearly define with direction is the positive direction.

Answer 1

Solution :- (a) mass of object, m=2 bound initial Velocity, V 6) = 2 feet is upwards air resistance, Fa= K Where K = 0.01 lb slft taking up the positive direction If this object is thrown upwards with initial velocity as given then initicil value problem will be mdo -mg - kv) and v(o)= 2 feet/s - K=0.01 e6 slft air resistance and weight my will be down words. so they are taken negative I dt (6) m=2 bound V(o)= 2 feet/s upwards K=0.01 lbs/ft If down is the positive direction -mov = mg + ku v (o)= a feet/s K=o.ol eb sift dt = mg tku 1 - - de 9+ Kve) weight and air resistence will act downward so they are to fien positive.

@ me Pkg VOO)=0 Fa= kv K=o0ool N. elm Let us take down as positive direction and up as negative direction. When an object is released from rest .suppose at a position in space that we take as origin, while falling downward, Net force more and weight mg will act down words hence taken positive and air Resistance will act upwards hence taken as negative Thus equation become dt mdo = mg -kv v(o)=0 m=aKg K=0.01 Nisim at иг negative positive Down