Suppose that the velocity vt in meters per second of a sky d

Suppose that the velocity

v(t)

(in meters per second) of a sky diver falling near the Earth\'s surface is given by the following exponential function, where time

t is the time after diving measured in seconds.

v(t)=71-71e^-0.19t

How many seconds after diving will the sky diver\'s velocity be

59

meters per second?

Round your answer to the nearest tenth, and do not round any intermediate computations.

Solution

v(t) = 71 -71(e^(-0.19t)

How many seconds after diving will the sky diver\'s velocity be 59 meters per second?

Plug v(t) = 59 m/sec

59 = 71 -71(e^(-0.19t)

-12 = -71(e^(-0.19t)

Divide both sides by 71 : 0.17 = e^(-0.19t)

Taking natural log of both sides: ln(0.17) = -0.19t

1.77 = 0.19t

t = 9.3 seconds