Suppose that the velocity [tex]\(v(t)\)[/tex] (in meters per second) of a sky diver falling near the Earth's surface is given by the following exponential function, where time [tex]\(t\)[/tex] is the time after diving measured in seconds.

[tex]\[v(t) = 61 - 61 e^{-0.26 t}\][/tex]

How many seconds after diving will the sky diver's velocity be 43 meters per second? Round your answer to the nearest tenth, and do not round any intermediate computations.

[tex]\[\boxed{\text{ seconds}}\][/tex]



Answer :

To find the time [tex]\( t \)[/tex] when the sky diver's velocity reaches 43 meters per second, given the velocity function [tex]\( v(t) = 61 - 61 e^{-0.26 t} \)[/tex], we need to set [tex]\( v(t) \)[/tex] equal to 43 and solve for [tex]\( t \)[/tex].

Start with the equation:
[tex]\[ v(t) = 43 \][/tex]

Substitute the given velocity function into this equation:
[tex]\[ 61 - 61 e^{-0.26 t} = 43 \][/tex]

Next, isolate the exponential term. Subtract 61 from both sides:
[tex]\[ 43 - 61 = -61 e^{-0.26 t} \][/tex]
[tex]\[ -18 = -61 e^{-0.26 t} \][/tex]

Divide both sides by -61 to solve for the exponential term:
[tex]\[ \frac{-18}{-61} = e^{-0.26 t} \][/tex]
[tex]\[ \frac{18}{61} = e^{-0.26 t} \][/tex]

Now, take the natural logarithm of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln\left(\frac{18}{61}\right) = -0.26 t \][/tex]

Solve for [tex]\( t \)[/tex] by dividing both sides of the equation by -0.26:
[tex]\[ t = \frac{\ln\left(\frac{18}{61}\right)}{-0.26} \][/tex]

At this point, calculate the expression using the exact value inside the logarithm before dividing:

[tex]\[ t \approx \frac{\ln(0.29508)}{-0.26} \][/tex]

This yields a numerical value:
[tex]\[ t \approx 4.694238870296718 \][/tex]

Round this value to the nearest tenth:
[tex]\[ t \approx 4.7 \][/tex]

Thus, the sky diver's velocity will be 43 meters per second approximately [tex]\( \boxed{4.7} \)[/tex] seconds after diving.