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Suppose that the velocity v() (in m/s) of a sky diver falling near the Earth's surface is given by the following function, where time t is measured inseconds.v (t) = 59(1 -ee-0.199)Find the initial velocity of the sky diver and the velocity after 5 seconds.Round your answers to the nearest whole number as necessary.

Suppose that the velocity v in ms of a sky diver falling near the Earths surface is given by the following function where time t is measured insecondsv t 591 ee class=

Respuesta :

To find the initial velocity (which is the velocity when t=0), substitute t=0 into the function v(t):

[tex]\begin{gathered} v(t)=59(1-e^{-0.19t}) \\ \Rightarrow v(0)=59(1-e^{-0.19\cdot0}) \\ =59(1-e^0) \\ =59(1-1) \\ =59(0) \\ =0 \end{gathered}[/tex]

To find the velocity after 5 seconds, substitute t=5:

[tex]\begin{gathered} v(5)=59(1-e^{-0.19\cdot5}) \\ =59(1-e^{-0.95}) \\ =59(1-0.38674\ldots) \\ =59(0.61326\ldots) \\ =36.18234\ldots \\ \approx36 \end{gathered}[/tex]

Therefore, the initial velocity is 0 m/s and the velocity after 5 seconds is 36 m/s.

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