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A skydiver is falling straight down, along the negative \(y\) direction. (a) During the initial part of the fall, her speed increases from 16 to \(28 \mathrm{m} / \mathrm{s}\) in \(1.5 \mathrm{s},\) as in part \(a\) of the figure. (b) Later, her parachute opens, and her speed decreases from 48 to \(26 \mathrm{m} / \mathrm{s}\) in \(11 \mathrm{s}\), as in part \(b\) of the drawing. Concepts: (i) Is her average acceleration positive or negative when her speed is increasing in part \(a\) of the figure? (ii) Is her average acceleration positive or negative when her speed is decreasing of the figure) determine the magnitude and direction of her average acceleration.

Short Answer

Expert verified
In part (a), acceleration is -8 m/s². In part (b), acceleration is +2 m/s².

Step by step solution

01

Understanding Acceleration

Acceleration is defined as the rate of change of velocity. It can be calculated using the formula \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time.
02

Calculate Acceleration (Part a)

In part (a), the skydiver's speed increases from 16 m/s to 28 m/s over 1.5 seconds. The change in velocity \( \Delta v = 28 - 16 = 12 \) m/s. Substituting into the acceleration formula, we get \( a = \frac{12}{1.5} = 8 \) m/s².
03

Direction of Acceleration (Part a)

Since the skydiver's speed is increasing in the negative \( y \) direction, the acceleration is also in the negative direction. Hence, the average acceleration is \(-8\) m/s².
04

Calculate Acceleration (Part b)

In part (b), her speed decreases from 48 m/s to 26 m/s over 11 seconds. The change in velocity \( \Delta v = 26 - 48 = -22 \) m/s. Substituting into the acceleration formula, we get \( a = \frac{-22}{11} = -2 \) m/s².
05

Direction of Acceleration (Part b)

The velocity is decreasing in the negative \( y \) direction, which means the actual acceleration, which opposes this decrease, is positive. Hence, the average acceleration is \(+2\) m/s².

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration
To grasp the concept of acceleration, it's important to understand that it measures how quickly the velocity of an object changes over time. Acceleration is a vector quantity, meaning it has both a magnitude and a direction. When an object's speed increases, it experiences positive acceleration. Conversely, when the object slows down, it encounters negative acceleration, also known as deceleration.

The basic formula to calculate acceleration is given by: \[ a = \frac{\Delta v}{\Delta t} \] where \( \Delta v \) represents the change in velocity, and \( \Delta t \) is the change in time. This formula allows us to determine how quickly a skydiver's speed changes as they fall. Whether the acceleration is positive or negative depends on the reference direction assigned to the movement, typically the direction of gravity for skydiving.
Velocity Change
Velocity change is a critical factor in determining acceleration. It is simply the difference between the final and initial velocities of an object. In our skydiver example, we witness two distinct scenarios that depict this change.

  • In Part (a), the skydiver's speed increases from \(16 \ \mathrm{m/s}\) to \(28 \ \mathrm{m/s}\). This results in a velocity change of \(12 \ \mathrm{m/s}\).
  • In Part (b), her speed decreases from \(48 \ \mathrm{m/s}\) to \(26 \ \mathrm{m/s}\), resulting in a velocity change of \(-22 \ \mathrm{m/s}\).

A change in velocity can be positive or negative, which indicates whether the skydiver is speeding up or slowing down. Understanding these principles helps us analyze the effects of forces such as gravity and air resistance during skydiving.
Skydiver Motion
The motion of a skydiver is fascinating, governed by gravitational forces and air resistance. When a skydiver leaps from a plane, they accelerate downwards due to gravity. Initially, their speed increases as they fall in a downward direction, which in physics terms, is often along the negative \(y\) direction.

  • For Part (a), the skydiver is accelerating as the velocity is increasing, resulting in a negative acceleration due to the chosen downward coordinate direction.
  • In Part (b), when the parachute is deployed, the speed decreases significantly. This deceleration indicates that the air resistance is now greater, acting upwards against the motion. Consequently, the average acceleration here becomes positive.

These phases illustrate how different forces influence a skydiver's acceleration, which is key to ensuring safety and control during their descent.
Average Acceleration Calculation
Calculating average acceleration is essential for understanding the changes in speed during skydiving. Applying the formula \( a = \frac{\Delta v}{\Delta t} \), we can determine how fast the skydiver's velocity changes during each phase of her jump.

  • In Part (a), with a velocity change of \(12 \ \mathrm{m/s}\) over \(1.5\) seconds, the average acceleration is \(-8 \ \mathrm{m/s^2}\). The negative sign indicates the acceleration vector is in the negative \(y\) direction.
  • For Part (b), the velocity changes by \(-22 \ \mathrm{m/s}\) in \(11\) seconds, giving an average acceleration of \(+2 \ \mathrm{m/s^2}\). This positive value tells us that the acceleration is now opposing the velocity decrease.

Recognizing these average accelerations helps in understanding skydiving dynamics, highlighting how forces act on the body and ensuring that physiological limits are respected for safety.

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Most popular questions from this chapter

A top-fuel dragster starts from rest and has a constant acceleration of \(40.0 \mathrm{m} / \mathrm{s}^{2}\). Concepts: (i) At time \(t\) the dragster has a certain velocity. Keeping in mind that the dragster starts from rest, when the time doubles to \(2 t\), does the velocity also double? (ii) When the time doubles to \(2 t,\) does the displacement of the dragster also double? Calculations: What are the (a) final velocities and (b) displacements of the dragster at the end of \(2.0 \mathrm{s}\) and at the end of twice this time, or \(4.0 \mathrm{s} ?\)

A locomotive is accelerating at \(1.6 \mathrm{m} / \mathrm{s}^{2} .\) It passes through a \(20.0-\mathrm{m}-\) wide erossing in a time of \(2.4 \mathrm{s}\). After the locomotive leaves the crossing, how much time is required until its speed reaches \(32 \mathrm{m} / \mathrm{s} ?\)

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A cement block accidentally falls from rest from the ledge of a \(53.0-\mathrm{m}\) -high building. When the block is \(14.0 \mathrm{m}\) above the ground, a man, \(2.00 \mathrm{m}\) tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?

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