Math  /  Algebra

QuestionSuppose that the velocity v(t)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 tt is the time after diving measured in seconds. v(t)=5555e0.26tv(t)=55-55 e^{-0.26 t}
How many seconds after diving will the sky diver's velocity be 41 meters per second? Round your answer to the nearest tenth, and do not round any intermediate computations. \square seconds

Studdy Solution

STEP 1

What is this asking? When will a skydiver, whose speed is changing according to a formula, reach a speed of 4141 meters per second? Watch out! Make sure to use the correct order of operations when solving for tt, and don't forget to round your final answer!

STEP 2

1. Set up the equation
2. Isolate the exponential
3. Solve for tt
4. Round and celebrate

STEP 3

We're given the velocity function v(t)=5555e0.26tv(t) = 55 - 55e^{-0.26t} and we want to find the time tt when the velocity is 4141 meters per second.
So, we **set** v(t)v(t) equal to 4141: 41=5555e0.26t41 = 55 - 55e^{-0.26t}

STEP 4

Let's **isolate** the exponential term.
First, we **subtract** 5555 from both sides of the equation: 4155=55e0.26t41 - 55 = -55e^{-0.26t} 14=55e0.26t-14 = -55e^{-0.26t}

STEP 5

Now, we **divide** both sides by 55-55: 1455=e0.26t\frac{-14}{-55} = e^{-0.26t} 1455=e0.26t\frac{14}{55} = e^{-0.26t}

STEP 6

To get tt out of the exponent, we can take the **natural logarithm** of both sides: ln(1455)=ln(e0.26t)\ln\left(\frac{14}{55}\right) = \ln\left(e^{-0.26t}\right)

STEP 7

Using the property ln(ex)=x\ln(e^x) = x, we **simplify** the right side: ln(1455)=0.26t\ln\left(\frac{14}{55}\right) = -0.26t

STEP 8

Finally, we **divide** both sides by 0.26-0.26 to **isolate** tt: t=ln(1455)0.26t = \frac{\ln\left(\frac{14}{55}\right)}{-0.26}

STEP 9

Now, we can **calculate** the value of tt: t=1.360960.26t = \frac{-1.36096 \dots}{-0.26} t5.23448t \approx 5.23448 \dots

STEP 10

The problem asks us to round to the nearest tenth.
Our **final answer** is approximately 5.25.2 seconds.

STEP 11

The skydiver's velocity will be 4141 meters per second approximately 5.25.2 seconds after diving.

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