A person is parachute jumping. During the time between when she leaps out of the

Question

A person is parachute jumping. During the time between when she leaps out of the plane and when she opens her chute, her

altitude is given by an equation of the form y = b - c ( t+ ke^ -t/k)

where

e

is the base of natural logarithms, and b, c, and k are constants. Because of air resistance, her velocity does not increase at a steady rate as it would for an object falling in vacuum.

(a) What units would b,c, and k have to have for the equation to make sense?

(b) Find the person's velocity, v , as a function of time. [You will need to use the chain rule, and the fact that d^ex)=dx=d^ex

(c) Use your answer from part (b) to get an interpretation of the constant c. [Hint:e^-x approaches large values for x]

(d) Find the person's acceleration,a, as a function of time.

(e) Use your answer from part (d) to show that if she waits long, enough to open her chute, her acceleration will become very small

Explanation / Answer

a)

b has units of m

c has units of m/s

k has units of s

b) v = dx/dt = -c - ck e^(-t/k)*-1/k = - c + c e^(-t/k)

c) as t-> infinity v-> -c

so c is terminal velocity

d) a = dv/dt = c e^(-t/k) *-1/k = -c/k e^(-t/k)

e) t-> infinity e^(-t/k) ->0

so a -> 0

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