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. v(t)=48(1 - e^{-0.25t}). find the velocity of the sky diver after 3 seconds and after 5 seconds. round your answers to the nearest whole number as necessary. velocity after 3 seconds: m/s velocity after 5 seconds: m/s

Explanation:

Step1: Substitute t = 3 into the formula

$v(3)=48(1 - e^{- 0.25\times3})$
$v(3)=48(1 - e^{-0.75})$
Since $e^{-0.75}\approx0.4724$, then $v(3)=48\times(1 - 0.4724)=48\times0.5276 = 25.3248\approx25$ m/s

Step2: Substitute t = 5 into the formula

$v(5)=48(1 - e^{- 0.25\times5})$
$v(5)=48(1 - e^{-1.25})$
Since $e^{-1.25}\approx0.2865$, then $v(5)=48\times(1 - 0.2865)=48\times0.7135 = 34.248\approx34$ m/s

Answer:

Velocity after 3 seconds: 25 m/s
Velocity after 5 seconds: 34 m/s