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Chapter 2: Problem 14

Review Conceptual Example 7 as background for this problem. A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is \(27.0 \mathrm{m} / \mathrm{s},\) and during a 5.0 \(\mathrm{s}\) - interval, it changes to a final speed of (a) 29.0 \(\mathrm{m} / \mathrm{s}\) and \((\mathrm{b}) 23.0 \mathrm{m} / \mathrm{s}\) . In each case, find the acceleration ( magnitude and algebraic sign) and state whether or not the car is decelerating.

Short Answer

Expert verified
Case (a): Acceleration is -0.4 m/s虏 (not decelerating). Case (b): Acceleration is 0.8 m/s虏 (decelerating).

Step by step solution

01

Understand the Problem

We need to calculate acceleration, defined as the change in velocity over time, for two scenarios. The car initially moves left with a velocity of \(-27.0\, \text{m/s}\). We will determine if the car is accelerating or decelerating in each case and identify the sign of the acceleration.
02

Use the Acceleration Formula for Case (a)

For this case, the final speed is \(-29.0\, \text{m/s}\). The formula for acceleration is: \[ a = \frac{v_f - v_i}{\Delta t} \] where \(v_f = -29.0\, \text{m/s}\), \(v_i = -27.0\, \text{m/s}\), and \(\Delta t = 5.0\, \text{s}\).
03

Calculate Acceleration for Case (a)

Substitute the values into the formula: \[ a = \frac{-29.0\, \text{m/s} - (-27.0\, \text{m/s})}{5.0\, \text{s}} = \frac{-2.0\, \text{m/s}}{5.0\, \text{s}} = -0.4\, \text{m/s}^2 \]
04

Interpret the Results for Case (a)

The negative sign indicates the acceleration is in the same direction as the velocity (left), but since the speed is increasing (more negative), the car is not decelerating.
05

Use the Acceleration Formula for Case (b)

In this case, the final speed is \(-23.0\, \text{m/s}\). Again use: \[ a = \frac{v_f - v_i}{\Delta t} \] with \(v_f = -23.0\, \text{m/s}\) and the other values unchanged.
06

Calculate Acceleration for Case (b)

Substitute into the formula: \[ a = \frac{-23.0\, \text{m/s} - (-27.0\, \text{m/s})}{5.0\, \text{s}} = \frac{4.0\, \text{m/s}}{5.0\, \text{s}} = 0.8\, \text{m/s}^2 \]
07

Interpret the Results for Case (b)

The positive acceleration indicates the direction is opposite to the velocity (still left), leading to a decrease in speed, hence the car is decelerating.

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Key Concepts

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

Understanding Acceleration
Acceleration is a fundamental concept in kinematics, describing how the speed of an object changes over time. It is a vector, meaning it has both magnitude and direction. In mathematical terms, acceleration is defined as the rate of change of velocity over time, represented by the formula \( a = \frac{\Delta v}{\Delta t} \).
When the final speed \( v_f \) is greater than the initial speed \( v_i \), acceleration has the same direction as the motion. Conversely, if \( v_f \) is less than \( v_i \), the acceleration opposes the motion.
  • A positive acceleration increases speed if moving in a positive direction.
  • A negative acceleration increases speed if moving in a negative direction.
  • A positive acceleration decreases speed if moving in a negative direction.
  • A negative acceleration decreases speed if moving in a positive direction.
Understanding these distinctions is key to interpreting whether an object is speeding up or slowing down.
Diving into Velocity
Velocity is another key concept in kinematics, indicating how fast a body is moving and in which direction. Unlike speed, which is a scalar, velocity is a vector and must always have a direction associated with it.
A change in velocity is what we call acceleration. In exercises like the one we discussed, initial and final velocities help determine whether a vehicle is speeding up or slowing down. If the distance covered per unit time changes, the velocity changes, and consequently, acceleration occurs.
  • Velocity to the left is often represented negatively, such as the car traveling left at \(-27.0\, \text{m/s}\).
  • Changes in velocity help in determining the type of acceleration involved.
  • Constant velocity means zero acceleration since speed and direction remain unchanged.
Grasping velocity helps clarify how objects move in any situation.
Deceleration Unpacked
Deceleration occurs when an object experiencing motion slows down, ultimately having negative acceleration concerning its direction of travel. It's important to understand that deceleration is not about moving backward, but rather, reducing the speed.
In the exercise, when the speed decreased from \(-27.0\, \text{m/s}\) to \(-23.0\, \text{m/s}\), the calculated acceleration was \(0.8\, \text{m/s}^2\). This positive value indicates that the acceleration works in the opposite direction, thus slowing the car.
  • Deceleration involves positive acceleration when moving in a negative direction.
  • It results in reduced speed and eventually stops the object if the force continues.
  • An understanding of deceleration aids in interpreting real-world motion, such as vehicles coming to a stop.
Deceleration aligns with practical situations such as braking.
Exploring Uniform Motion
Uniform motion describes when an object moves constantly either at a steady speed or in a straight line. In such cases, velocity does not change, which implies zero acceleration.
When a car continues to move left with a uniform motion of \(-27.0\, \text{m/s}\), it maintains its velocity without speeding up or slowing down.
  • Uniform motion implies a steady, unchanged state of velocity.
  • It negates any presence of acceleration, as there's no change in speed or direction.
  • Whenever forces are balanced, uniform motion is a likely outcome.
Understanding uniform motion helps in predicting an object's path and is fundamental in mechanics, allowing for simple motion descriptions.

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Most popular questions from this chapter

A car is traveling at a constant speed of 33 \(\mathrm{m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 2.5 \(\mathrm{km}\) away?

(a) Suppose that a NASCAR race car is moving to the right with a constant velocity of \(+82 \mathrm{m} / \mathrm{s}\) . What is the average acceleration of the car? (b) Twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed. What is the average acceleration of the car?

(a) What is the magnitude of the average acceleration of a skier who, starting from reaches a speed of 8.0 \(\mathrm{m} / \mathrm{s}\) when going down a slope for 5.0 \(\mathrm{s} ? \quad\) (b) How far does the skier travel in this time?

An Australian emu is running due north in a straight line at a speed of 13.0 \(\mathrm{m} / \mathrm{s}\) and slows down to a speed of 10.6 \(\mathrm{m} / \mathrm{s}\) in 4.0 s. (a) What is the direction of the bird's acceleration? (b) Assuming that the acceleration remain the same, what is the bird's velocity after an additional 2.0 \(\mathrm{s}\) has elapsed?

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 \(\mathrm{m} / \mathrm{s}\) . The car is a distance \(d\) away. The bear is 26 \(\mathrm{m}\) behind the tourist and running at 6.0 \(\mathrm{m} / \mathrm{s}\) . The tourist reaches the car safely. What is the maximum possible value for \(d ?\)
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