Question

Suppose that the velocity v(t) (in m/s) of a sky diver falling near the Earth’s surface is given by the following function, where time t is measured in seconds.

Answer 1

The expression is given as,

`v(t)=53(1-e^(-0.26t))`

Substitute the value t=3 in the above expression and obtain the velocity corresponding to 3 seconds,

`\begin{gathered} v(3)=53(1-e^(-0.26(3))) \\ v(3)=53(1-e^(-(0.78))) \\ v(3)\approx53(0.54159) \\ v(3)\approx28.7045 \\ v(3)\approx29 \end{gathered}`

Thus, the velocity of the skydiver is 29 m/s after 3 seconds.

Substitute the value t=5 in the above expression and obtain the velocity corresponding to 5 seconds,

`\begin{gathered} v(5)=53(1-e^(-0.26(5))) \\ v(5)=53(1-e^(-1.3)) \\ v(5)\approx53(0.727) \\ v(5)\approx38.5558 \\ v(5)\approx39 \end{gathered}`

Thus, the velocity of the skydiver is 39 m/s approximately after 5 seconds.