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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.

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 tim class=

Answer :

The equation of the velocity is given by the exponential:

[tex]v(t)=53-53e^{-0.24t}[/tex]

Let us say that the sky driver's velocity will be 47 m/s at t₁. Then, using the expression above:

[tex]\begin{gathered} v(t_1)=47 \\ 53-53e^{-0.24t_1}=47 \end{gathered}[/tex]

Solving for t₁:

[tex]\begin{gathered} \frac{53-47}{53}=e^{-0.24t_1} \\ \ln (\frac{6}{53})=-0.24t_1 \\ t_1=9.1s \end{gathered}[/tex]

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