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Chapter 3: Problem 39

A skydiver is falling straight down at \(55 \mathrm{m} / \mathrm{s}\) when he opens his parachute and slows to \(8.3 \mathrm{m} / \mathrm{s}\) in $3.5 \mathrm{s} .$ What is the average acceleration of the skydiver during those 3.5 s?

Short Answer

Expert verified
Answer: The skydiver's average acceleration during these 3.5 seconds is -13.34 m/s^2.

Step by step solution

01

Identify the given information

The problem gives us the following information: - Initial velocity, \(v_i = 55 \mathrm{m} / \mathrm{s}\) - Final velocity, \(v_f = 8.3 \mathrm{m} / \mathrm{s}\) - Time, \(t = 3.5 \mathrm{s}\)
02

Write down the formula for average acceleration

The formula to calculate average acceleration is given by: Average acceleration = \(\frac{v_f - v_i}{t}\)
03

Substitute the values into the formula

Now, we will plug the given values into the formula: Average acceleration = \(\frac{8.3\,\mathrm{m/s} - 55\,\mathrm{m/s}}{3.5\,\mathrm{s}}\)
04

Calculate the average acceleration

Carry out the calculations: Average acceleration = \(\frac{-46.7\,\mathrm{m/s}}{3.5\,\mathrm{s}} = -13.34\,\mathrm{m/s^2}\)
05

Interpret the result

In this case, the average acceleration during the 3.5 seconds is \(-13.34 \,\mathrm{m/s^2}\). The negative sign indicates that the skydiver is slowing down as he experiences upwards (deceleration) force due to opening of parachute.

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