/*! 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 4 A particle is moving in \(x-y\) ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 4

A particle is moving in \(x-y\) plane. At time \(t=0\), particle is at \((1 \mathrm{~m}, 2 \mathrm{~m})\) and has velocity \((4 \hat{i}+6 \hat{j}) \mathrm{m} / \mathrm{s}\). Al \(t=4 \mathrm{~s}\), particle reaches at \((6 \mathrm{~m}, 4 \mathrm{~m})\) and has velocity \((2 \hat{i}+10 \hat{j}) \mathrm{m} / \mathrm{s}\). In the given time interval, find (a) average velocity (b) average acecleration and (c) from the given data, can you find avcrage spoed?

Short Answer

Expert verified
(a) Average velocity: \(1.25 \hat{i} + 0.5 \hat{j} \ \mathrm{m/s}\). (b) Average acceleration: \(-0.5 \hat{i} + 1 \hat{j} \ \mathrm{m/s}^2\). (c) Average speed cannot be determined.

Step by step solution

01

Calculate Displacement

Displacement is the change in position of the particle from its initial position to its final position. The initial position is \( (1 \mathrm{~m}, 2 \mathrm{~m}) \) and the final position is \((6 \mathrm{~m}, 4 \mathrm{~m})\).\ The displacement \( \vec{d} \) is given by: \[ \vec{d} = (6 - 1) \hat{i} + (4 - 2) \hat{j} = 5 \hat{i} + 2 \hat{j} \ \mathrm{m} \] This is the vector displacement of the particle over the time interval.
02

Calculate Average Velocity

The average velocity \( \vec{v}_{\text{avg}} \) is the total displacement divided by the time interval. The displacement vector \( \vec{d} = 5 \hat{i} + 2 \hat{j} \ \mathrm{m} \), and the time interval is from \(t = 0\) to \(t = 4\) seconds, thus \(\Delta t = 4\) s.The formula for average velocity is: \[ \vec{v}_{\text{avg}} = \frac{\vec{d}}{\Delta t} = \frac{5 \hat{i} + 2 \hat{j}}{4} = 1.25 \hat{i} + 0.5 \hat{j} \ \mathrm{m/s} \] This is the average velocity over the interval.
03

Calculate Initial and Final Velocities

The initial velocity \( \vec{u} \) is given by \( (4 \hat{i} + 6 \hat{j}) \ \mathrm{m/s} \) and the final velocity \( \vec{v} \) is given by \( (2 \hat{i} + 10 \hat{j}) \ \mathrm{m/s} \). These values are provided and will be used to calculate the average acceleration.
04

Calculate Average Acceleration

The average acceleration \( \vec{a}_{\text{avg}} \) is the change in velocity divided by the time interval. Use the initial velocity \( \vec{u} = 4 \hat{i} + 6 \hat{j} \ \mathrm{m/s} \) and final velocity \( \vec{v} = 2 \hat{i} + 10 \hat{j} \ \mathrm{m/s} \). The formula for average acceleration is: \[ \vec{a}_{\text{avg}} = \frac{\vec{v} - \vec{u}}{\Delta t} = \frac{(2 \hat{i} + 10 \hat{j}) - (4 \hat{i} + 6 \hat{j})}{4} \]\[ = \frac{-2 \hat{i} + 4 \hat{j}}{4} = -0.5 \hat{i} + 1 \hat{j} \ \mathrm{m/s}^2 \] This is the average acceleration over the interval.
05

Analyze Average Speed

Average speed is the total distance traveled divided by the time interval. With only initial and final positions and velocities, the exact path of motion is unknown, so the distance traveled is unknown. Thus, we cannot determine the average speed from the given data.

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Key Concepts

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

Displacement
Displacement in motion describes the change of position of an object. It is not the total path taken but the shortest straight-line distance from the starting point to the ending point. In this exercise, a particle moves from an initial position of \((1 \mathrm{~m}, 2 \mathrm{~m})\) to a final position of \((6 \mathrm{~m}, 4 \mathrm{~m})\). The formula to calculate displacement is:
  • Subtract the initial coordinates from the final coordinates.
  • Displacement vector \( \vec{d} = (\text{final} - \text{initial}) \).
  • In this example: \( \vec{d} = (6 - 1) \hat{i} + (4 - 2) \hat{j} = 5 \hat{i} + 2 \hat{j} \ \mathrm{m} \).
Displacement is vector-based, which means both the magnitude and the direction are important in summarizing how far and which way an object has moved from its original point.
Average Velocity
Average velocity tells us the overall change in position over the change in time. It takes into account the direction of motion, being a vector quantity like displacement.
  • Calculated by dividing the total displacement by the total time interval.
  • Formula: \[ \vec{v}_{\text{avg}} = \frac{\vec{d}}{\Delta t} \]
  • In our scenario, using displacement \(5 \hat{i} + 2 \hat{j} \ \mathrm{m}\) and time interval \(4 \ \text{s}\), the average velocity becomes: \( \frac{5 \hat{i} + 2 \hat{j}}{4} = 1.25 \hat{i} + 0.5 \hat{j} \ \mathrm{m/s} \).
Average velocity is pivotal in understanding how fast and in what direction an object is moving overall, as it smooths out any fluctuations in speed and direction during the covered time.
Average Acceleration
Average acceleration provides information about how the velocity of an object changes over time. Unlike velocity, acceleration factors in changes to speed as well as changes in the direction of motion.
  • It鈥檚 computed by dividing the change in velocity by the time interval.
  • The change in velocity is determined by subtracting the initial velocity \( \vec{u} \) from the final velocity \( \vec{v} \).
  • Formula: \[ \vec{a}_{\text{avg}} = \frac{\vec{v} - \vec{u}}{\Delta t} \]
  • In this case, with initial velocity \( (4 \hat{i} + 6 \hat{j}) \ \mathrm{m/s} \) and final velocity \( (2 \hat{i} + 10 \hat{j}) \ \mathrm{m/s} \), the change in velocity is \((-2 \hat{i} + 4 \hat{j})\) and,as it spans \(4 \ \text{s}\), the average acceleration results in: \( -0.5 \hat{i} + 1 \hat{j} \ \mathrm{m/s}^2 \).
Average acceleration reveals how velocity vectors evolve over time and can highlight changes due to forces acting upon the object.

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Most popular questions from this chapter

A particle moves in a circle of radius \(R=\operatorname{lm}\) such that its spced varics with time as \(v=0.5 e^{t} \mathrm{~m} / \mathrm{s}\). Find the initial acccleration of the particle.

\(\Lambda\) particle moves in a circle of radius \(25 \mathrm{~cm}\) at two revolutions per second 'lhe acceleration of the particle in \(\mathrm{m} / \mathrm{s}^{2}\) is (a) \(\pi^{2}\) (b) \(8 \pi^{2}\) (c) \(4 \pi^{2}\) (d) \(2 \pi^{2}\)

A particle is to be projected so as to just pass through three equal rings of diameter \(d\) and placed in parallel vertical planes at distances \(a\) apart with their highest points at a height \(h\) above the point of projection. Prove that the clcvation of projection is tan \(\frac{\sqrt{h d}}{a}\).

Statement-1: The average velocity of a body in a general projectilc motion during any time interval \(A \leq T\), is uniform, if the body is at the same height above the ground level at two extremities of the time interval \(A t . T\) is the flight time o the projectile Statement-2: 'lhere is no acceleration acting on the projectile in horizontal direction.

From a point at a height \(h\) above the oround, a particle \(A\) is projected with a velocity \(v\) in an upward direction making an angle \(\theta\) with the horizontal. \(\Lambda\) nother particle \(B\) is projected from the same point with the same velocity \(v\) but, in a direction directly opposite to \(A\). Show that the two particles hit the ground al a distanec \(\frac{2 v}{g} \cos \theta \sqrt{\left(v^{2} \sin ^{2} \theta+2 g h\right)}\) apart.
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