Question

The equation describing the velocity of the ice falling from the wing as a function of time is v=45(1-(0.804)^t). Determine algebraically the time required, to the nearest tenth of a second, for the ice to reach a velocity of 98% of its terminal velocity of 45 m/s.

(v=velocity)

Please help me. Thank-you!

Answer 1

Since it is not specified in the question, we will assume that the equation for v as a function of time is in m/s.
Thus
v(t)=45(1-(0.804)^t) m/s
where t is in seconds.
A plot of the graph of v(t) versus t will shed some light.
http://i263.photobucket.com/albums/ii157/mathmate/terminalVelocity.png

Let v1=98% of terminal velocity of 45 m/s
then we look for the value of t1 such that
v(t1)=45*0.98
or
45*(1-0.804^t1) = 45*0.98
0.804^t1=1-0.98=0.02
Can you take it from here?
Hint: solve for t1 either by trial and error or apply the laws of logarithm.

Answer 2

0.804^t= 0.02
t*log(0.804)-log(0.02)
t=log(0.02)/ log(0.804)
t= 17.9

Answer 3

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