/*! 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 1 The \(x\) co-ordinate of a parti... [FREE SOLUTION] | 91Ó°ÊÓ

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The \(x\) co-ordinate of a particle moving along \(x\) -axis in terms of time is given as \(x=(2-t)(t-6)\) metre. 'The displacement in time \(t=2\) to \(t=5 \mathrm{sec}\) is (a) \(2 \mathrm{~m}\) (b) \(3 \mathrm{~m}\) (c) \(4 \mathrm{~m}\) (d) None

Short Answer

Expert verified
The displacement is 3 m.

Step by step solution

01

Identify the Initial and Final x-coordinates

To find the displacement, we need to calculate the position of the particle at the starting and ending times. First, we identify that we need the x-coordinates at \( t = 2 \) and \( t = 5 \). The x-coordinate function is given by \( x = (2-t)(t-6) \).
02

Calculate x-coordinate at t = 2

Substitute \( t = 2 \) into the given function: \[x = (2-2)(2-6) = 0 \]This gives us the position of the particle at \( t = 2 \).
03

Calculate x-coordinate at t = 5

Substitute \( t = 5 \) into the given function: \[x = (2-5)(5-6) = (-3)(-1) = 3\]This gives us the position of the particle at \( t = 5 \).
04

Calculate the Displacement

The displacement is found by taking the difference between the final position and the initial position: \[\text{Displacement} = x(t = 5) - x(t = 2) = 3 - 0 = 3 \]Thus, the displacement of the particle from \( t = 2 \) to \( t = 5 \) is \( 3 \) meters.

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Key Concepts

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

Displacement in Physics
Displacement is a fundamental concept in physics, especially when analyzing motion. It refers to the change in position of an object from one point to another and is a vector quantity. This means it has both magnitude and direction. Unlike distance, displacement is concerned with the overall change in position and not the path taken. For example, if an object moves 3 meters to the right from its starting point, its displacement is simply 3 meters to the right, regardless of how it got there.
In mathematical terms, displacement is calculated using the formula:
\[ \text{Displacement} = x_f - x_i \]
where \(x_f\) is the final position and \(x_i\) is the initial position. In the context of the exercise, the displacement from \(t = 2\) to \(t = 5\) is calculated as:
\[ \text{Displacement} = 3 \text{ meters} - 0 \text{ meters} = 3 \text{ meters} \]
This calculation shows how the particle has moved along the \(x\)-axis over a specific time period.
Kinematics in One Dimension
Kinematics is the study of motion without considering the forces causing it. When dealing with motion in one dimension, we analyze how objects move in a straight line. This simplifies problems by reducing them to movements along a single axis.
In one-dimensional kinematics, we often use equations that relate the position, velocity, and acceleration of an object. For constant acceleration, these kinematic equations come into play, but in simpler scenarios like the one in the exercise, we use position functions.
By understanding the motion along one dimension, we can solve various problems relating to how objects move from one point to another over time. The position of the particle at any time \(t\) can be calculated using the given function \( x = (2-t)(t-6) \), which is a key aspect of one-dimensional motion analysis.
Motion Along X-Axis
Studying motion along the \(x\)-axis is an example of one-dimensional motion. It involves understanding how an object moves either positively or negatively along a horizontal line. This can be referred to as linear motion and is one of the simplest forms of motion to analyze.
Motion along the \(x\)-axis can be defined using a function of time as seen in the exercise with \(x = (2-t)(t-6)\). Here, the position of the particle can be determined for any time \(t\).
  • An object moving to the right or upward is considered to have positive displacement or motion.
  • Conversely, movement to the left or downward indicates negative displacement.

Understanding motion along an axis is critical not just in theoretical physics, but also in practical applications like engineering and robotics design, where precise measurements and calculations of linear motion are required.

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Most popular questions from this chapter

A body is moving with uniform acceleration starting from rest. Its motion is deseribed by some interpreted results. Column-I 1 (p) Column-II \(v_{1}: v_{2} \quad v_{3}=1: 2: 3\) (a) Its velocity at the end \(o^{2} t\) sec., \(2 t\) sec. \(3 t \mathrm{sec}\). are in the ratio (b) Distances covered in \(\mathrm{I}^{* 1} t\) sec., \(2^{\text {all }} t\) sec. (q) \(v_{1}: v_{2} \quad v_{3}=1: 1: 1\) \(3^{\text {ra }} t\) sec. are in the ratio (c) Distances covered in \(t\) sec. \(2 t\) see. (r) \(s_{1}: s_{2}: s_{3}=1: 3: 5\) \(3 t\) sec. are in the ratio (d) Change in velocity at the cnd of \(1^{A}\) (s) \(s_{1}: s_{2}: s_{3}=1: 4: 9\) \(t\) sec. \(2^{\text {nd }} t \sec , 3^{\text {nis }} t\) sec. are in the ratio

'Ihe speed of a body moving on a straight track varies according to \(v=2 t+13\) for \(0 \leq t \leq 5 \mathrm{~s}\), \(v=3 t+8\) for \(57 \mathrm{~s}\). The distance moved by the particle at the end of 10 second is: (a) \(127 \mathrm{~m}\) (b) \(247 \mathrm{~m}\) (c) \(186 \mathrm{~m}\) (d) \(3133 \mathrm{~m}\)

A marble ball, after having fallen from rest undor the influenee of gravity for \(6 \mathrm{sec}\). erashes through a horizontal glass plate, thercby losing two-third of its velocity. If it then reaches the ground in \(2 \mathrm{sec}\), find the height of the plate above the ground

The velocity ol a particle moving in a straight line varies with time in such a manner that \(v\) versus \(t\) graph is represented by one hal o \([\) an circle. The maximum velocity is \(v_{m}\) and total time of motion is \(t_{0}\) (i) average velocity of particle is \(\frac{\pi}{4} v_{n 1}\) (ii) such molion cannot be realized in praclical terms (a) only (i) is correct (b) only (ii) is correct (c) both (i) and (ii) are correct (d) both (i) and (ii) are wrong.

Velocily-time cquation of a particle moving in a straight line is, \(v=\left(10+2 t+3 t^{2}\right) \mathrm{m} / \mathrm{s}\). Find, (a) displacement of particle from the fixed position at time \(t=1 \mathrm{~s}\), il \(\mathrm{it}\) is given that displacement is \(20 \mathrm{~m}\) at time \(t=0\) (b) acecleration-time cquation.

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