/*! 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 131 The displacement of a particle i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 131

The displacement of a particle is given by \(x=(t-2)^{2}\), where \(x\) is in metres and \(t\) in seconds. The distance covered by the particle in first \(4 \mathrm{~s}\) is (A) \(4 \mathrm{~m}\) (B) \(8 \mathrm{~m}\) (C) \(12 \mathrm{~m}\) (D) \(16 \mathrm{~m}\)

Short Answer

Expert verified
The distance covered by the particle in the first 4 seconds is \(8\mathrm{~m}\).

Step by step solution

01

Identify the Displacement Function

The displacement function is given by \(x = (t - 2)^2\). This formula gives the position of the particle at any time t.
02

Calculate the Initial Position

To find the initial position of the particle, we need to evaluate the displacement function at time \(t=0\): \[x(0) = (0 - 2)^2 = 4\]
03

Calculate the Position at t=4 seconds

To find the position of the particle at \(t=4\mathrm{~s}\), we need to evaluate the displacement function at time \(t=4\): \[x(4) = (4 - 2)^2 = 2^2 = 4\]
04

Calculate the Distance Covered

The distance covered by the particle in the first 4 seconds is the difference between its positions at \(t=4\mathrm{~s}\) and \(t=0\): \[Distance = x(4) - x(0) = 4 - 4 = 0\] However, since distance should be a positive value, we should consider the motion of the particle during the given time interval. To find the range of motion, we need to find the turning points of the function within the given interval. A turning point occurs when the velocity changes its sign, and that happens when the derivative of the function is equal to zero.
05

Calculate the Derivative of the Function

To calculate the derivative of the displacement function, we differentiate with respect to t: \[\frac{dx}{dt} = \frac{d(t-2)^2}{dt} = 2(t-2)\]
06

Find the Turning Points

To find the turning point, we can set the derivative equal to zero and solve for t: \[2(t-2) = 0\] \[t = 2\] So there is a turning point at t=2 seconds.
07

Calculate the Distance from t=0 to t=2 seconds

Now we need to find the displacement from t=0 to t=2 seconds using the displacement function: \[x(2) = (2 - 2)^2 = 0\] The distance from t=0 to t=2 is equal to: \[Distance_{0-2} = x(0) - x(2) = 4-0 = 4\mathrm{~m}\]
08

Calculate the Distance from t=2 to t=4 seconds

We already calculated the displacement at t=4 seconds. So the distance from t=2 to t=4 seconds is equal to: \[Distance_{2-4} = x(4) - x(2) = 4-0 = 4\mathrm{~m}\]
09

Calculate the Total Distance Covered

The total distance covered by the particle in the first 4 seconds is the sum of the distances calculated in Steps 7 and 8: \[Total\, Distance = Distance_{0-2} + Distance_{2-4} = 4\mathrm{~m} + 4\mathrm{~m} = 8\mathrm{~m}\] The distance covered by the particle in the first 4 seconds is \(8\mathrm{~m}\), which corresponds to option (B).

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!

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

A vehicle travels half the distance \(L\) with speed \(V_{1}\) and the other half with speed \(V_{2}\), then its average speed is (A) \(\frac{V_{1}+V_{2}}{2}\) (B) \(\frac{2 V_{1}+V_{2}}{V_{1}+V_{2}}\) (C) \(\frac{2 V_{1} V_{2}}{V_{1}+V_{2}}\) (D) \(\frac{2 V_{1} V_{2}}{V_{1}+V_{2}}\)

The velocity \(v\) of a body moving along a straight line is varying with time \(t\) as \(v=t^{2}-4 t\), where \(v\) in \(\mathrm{m} / \mathrm{s}\) and \(t\) in seconds. The magnitude of initial acceleration is (A) Zero (B) \(2 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(4 \mathrm{~m} / \mathrm{s}^{2}\) (D) \(6 \mathrm{~m} / \mathrm{s}^{2}\)

The position \((x)\) of a particle varies with time as \(x=\) \(t^{3}-3 t^{2}\), where \(t\) is time in second. Match Column-I with Column-II. $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (A) Position of particle is } & \text { (1) } t=3 \mathrm{~s} \\\ \text { zero at } \\ \text { (B) Velocity of particle is } & \text { (2) } t=2 \mathrm{~s} \\ \text { zero at } \\ \text { (C) Magnitude of } & \text { (3) } t=5 \mathrm{~s} \\ \text { acceleration of particle } \\ \text { is zero at } \\ \begin{array}{ll} \text { (D) Magnitude of } \\ \text { instantaneous } \\ \text { acceleration is equal to } \\ 18 \mathrm{~m} / \mathrm{s}^{2} \end{array} & \text { (4) } t=1 \mathrm{~s} \\ \hline \end{array} $$

A ball whose kinetic energy is \(E\) is projected at an angle of \(45^{\circ}\) to the horizontal. The kinetic energy of the ball at the highest point of its flight will be \([\mathbf{2 0 0 2}]\) (A) \(E\) (B) \(E / \sqrt{2}\) (C) \(E / 2\) (D) Zero

Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity \(v\) and other with a uniform acceleration \(a\). If \(\alpha\) is the angle between the lines of motion of two particles then the least value of magnitude of relative velocity will be at time given by (A) \(\frac{v}{a} \sin \alpha\) (B) \(\frac{v}{a} \cos \alpha\) (C) \(\frac{v}{a} \tan \alpha\) (D) \(\frac{v}{a} \cot \alpha\)
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Electromagnetism

Read Explanation

Radiation

Read Explanation

Astrophysics

Read Explanation

Space Physics

Read Explanation

Solid State 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.