Question

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.

Answer 1

The equation of the velocity is given by the exponential:

`v(t)=53-53e^(-0.24t)`

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

`\begin{gathered} v(t_1)=47 \\ 53-53e^(-0.24t_1)=47 \end{gathered}`

Solving for t₁:

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