/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Statement-1 : A negative acceler... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 22

Statement-1 : A negative acceleration of a body is associated with a slowing down of body Statement-2 : Acceleration is a vector quantity.

Short Answer

Expert verified
Both statements are true; Statement-2 supports Statement-1, but the context of acceleration's direction is crucial.

Step by step solution

01

Analyze Statement-1

Statement-1 claims that negative acceleration is associated with slowing down. This is true if the direction of the acceleration is opposite to the direction of the velocity. However, if acceleration occurs in the same direction as velocity, it could actually speed up the object, just in a negative or opposite direction from the original motion.
02

Analyze Statement-2

Statement-2 states that acceleration is a vector quantity, which means it has both magnitude and direction. This is a factual definition of acceleration in physics, so Statement-2 is true.
03

Evaluate the Relationship Between the Statements

Statement-1's interpretation depends on the directional aspect of acceleration, fitting with Statement-2's assertion that acceleration is a vector. Thus, while Statement-2 gives the reasoning behind Statement-1, Statement-1 might be misleading if the vector direction isn't considered carefully.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Negative acceleration
Negative acceleration is often misunderstood. It is commonly associated with slowing down, but this isn't always the case. In physics, acceleration refers to any change in velocity, which can include speeding up, slowing down, or changing direction. Negative acceleration specifically refers to the direction of the acceleration vector. If an object's velocity is positive and its acceleration is negative, this could lead to the object slowing down. However, if both velocity and negative acceleration point in the same direction, the object might speed up in that direction instead.
It's crucial to consider the direction of the velocity and acceleration vectors. Just because acceleration is labeled "negative," it doesn't always imply deceleration. A negative value simply indicates direction within the right coordinate system. Hence, understanding the motion requires knowing both the direction and magnitude of acceleration.
Vector quantity
In physics, a vector quantity is one that possesses both magnitude and direction. Unlike scalar quantities, which have only magnitude, vectors are crucial for describing quantities like velocity, force, and acceleration. Acceleration, in particular, is a vector quantity. This means it requires both a value for how much the speed will change and a specific direction in which this change occurs.
Vector quantities are typically represented with arrows. The length of the arrow indicates the magnitude, while the arrow's direction shows the vector's direction. This representation helps one visualize the physical situation easily.
  • For example, if a car accelerates faster towards the east, the acceleration is a vector pointing east.
  • As another scenario, if the car slows down moving west, the vector points west even if the magnitude is labeled as negative.
This understanding is fundamental because the direction of the vector influences how we interpret the motion of objects.
Velocity and acceleration direction
The relationship between velocity and acceleration directions is pivotal to understanding motion. Velocity describes how fast and in what direction an object is moving. Meanwhile, acceleration describes how the velocity changes over time, accounting for changes in speed or direction.
When acceleration and velocity are in the same direction, the object speeds up. The result is increased velocity in the same vector direction. Conversely, if they are in opposite directions, the object slows down, leading to reduced velocity, even if it proceeds in the original direction.
  • If a car moves north with increasing speed, both velocity and acceleration vectors point north.
  • If the car moves north but decelerates, the acceleration vector points south.
Clearly, analyzing the directionality of vectors is crucial in predicting how an object's motion will evolve, enabling a comprehensive understanding of dynamics in physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The acceleration \(a\) in \(\mathrm{m} / \mathrm{s}^{2}\) of a particle is given by \(a=3 t^{2}+2 t+2\) where \(t\) is the time in sec. If the particle starts out with a velocity \(v=2 \mathrm{~m} / \mathrm{s}\) at \(t=0\), then find the velocity at the end of \(2 \mathrm{~s}\).

The position of a particle at any instant \(t\) is given by \(x=t^{2}-7 t\). The average velocity of the partielc from \(t=t_{1}\) to \(t=t_{2}\) is zero. Here \(t_{1}\), and \(t_{2}\) respectively, cannot be cqual to: (a) 1,6 (b) \(2.5\) (c) \(2.5,3.5\) (d) 0,7

The velocity of a particle moving in the positive direction o \([x\) -axis varies as \(v-10 \sqrt{x}\). Assuming that at \(t=0\) particle was at \(x=0\). Then (a) The initial velocity of the particle is zero (b) 'The initial velocity of the particle is \(2.5 \mathrm{~m} / \mathrm{s}\) (c) The acceleration of the particle is \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) (d) 'The acceleration of the particle is \(50 \mathrm{~m} / \mathrm{s}^{2}\)

Suppose that a particle moves on a straight line so that its velocity at time \(t\) is \(v=\left(t^{2}-2 t\right) \mathrm{m} / \mathrm{s}\) (a) Find the displacement of the particle during the time interval \(0 \leq t \leq 3\). (b) Find the distance travelled by the particle during the time interval \(0 \leq t \leq 3\).

The position \(y\) of a particle depends on time as \(y=A \sin (\omega t+\varphi)\), where \(A, \omega\) and \(\varphi\) is a \((+v e)\) constant. At \(t=0\), the particle is at \(y-\frac{A}{2}\). Find \(\left(\frac{d^{2} y}{d t^{2}}\right)\) at \(t-\left(\frac{\pi}{\omega}\right) \mathrm{sec}\). (a) \(\frac{\omega^{2} A}{2}\) (b) \(\omega^{2} A\) (c) \(\frac{\omega^{2} A \sqrt{3}}{2}\) (d) zero
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Particle Model of Matter

Read Explanation

Dynamics

Read Explanation

Electromagnetism

Read Explanation

Torque and Rotational Motion

Read Explanation

Fields in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.