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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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Most popular questions from this chapter

At \(t=0,\) an automobile traveling north begins to make a turn. It follows one- quarter of the arc of a circle with a radius of \(10.0 \mathrm{m}\) until, at \(t=1.60 \mathrm{s},\) it is traveling east. The car does not alter its speed during the turn. Find (a) the car's speed, (b) the change in its velocity during the turn, and (c) its average acceleration during the turn.
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