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)=52(1-e^-0.21t)

Find the initial velocity of the sky diver and the velocity after 6 seconds.

Answer 1

Answer: Initial velocity = 0 m/s, velocity after 6 seconds = 37.44 m/s.

Step-by-step explanation:

Since we have given that

$v(t)=52(1-e^{-0.21t})$

We need to find the initial velocity.

As we know that initial velocity is at t = 0.

so, it becomes,

$v(0)=52(1-e^{-0.21\times 0})\\\\v(0)=0\ m/s$

Velocity after 6 seconds.

So, we put t = 6 seconds,

$v(6)=52(1-e^{-0.21\times 6})\\\\v(6)=52(1-e^{-1.26})\\\\v(6)=52\times 0.72\\\\v(6)=37.44\ m/s$

Hence, Initial velocity = 0 m/s, velocity after 6 seconds = 37.44 m/s.