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. v(t)=53−53e−0.23t How many seconds after diving will the sky diver's velocity be 36 meters per second? Round your answer to the nearest tenth, and do not round any intermediate computations.

Answer 1

7.7 seconds.

Step-by-step explanation:

To find the number of seconds after diving when the sky diver's velocity is 36 meters per second, we need to set the given velocity function equal to 36 and solve for t.

The given velocity function is v(t) = 53 - 53e^(-0.23t), where t is the time after diving in seconds.

Setting v(t) = 36, we have:

36 = 53 - 53e^(-0.23t)

To solve this equation for t, we need to isolate the exponential term on one side of the equation.

First, subtract 53 from both sides:

36 - 53 = -53e^(-0.23t)

Simplifying the left side:

-17 = -53e^(-0.23t)

Next, divide both sides by -53:

-17/-53 = e^(-0.23t)

Simplifying further:

0.3208 = e^(-0.23t)

To solve for t, we need to take the natural logarithm (ln) of both sides:

ln(0.3208) = ln(e^(-0.23t))

Using the property of logarithms that ln(e^x) = x:

ln(0.3208) = -0.23t

Finally, divide both sides by -0.23 to solve for t:

t = ln(0.3208) / -0.23

Using a calculator, we find that t ≈ 7.7356 seconds.

Rounding to the nearest tenth, the sky diver's velocity will be 36 meters per second approximately 7.7 seconds after diving.