
A particle moves in a straight line and its speed depends on time as $v = \left| {2t - 3} \right|$. Find the displacement of the particle in $5s$.
Answer
216.3k+ views
Hint: Recall that the displacement of a particle is defined as the shortest distance that is travelled by the particle between two points. The displacement of the particle depends on the initial and final position, and it is not necessary that the initial position should be at origin. Displacement of a particle is denoted by the symbol ‘s’ and can only be measured along a straight path. Also the displacement of a particle is always indicated by an arrow, where the arrow points in the direction of motion of the particle.
Complete step by step solution:
Given that the speed depends as $v = \left| {2t - 3} \right|$
Therefore, it can be written that
$\Rightarrow v = \left\{ {2t - 3,t \geqslant \dfrac{3}{2}} \right\}$
Or $v = \left\{ { - 2t + 3,t < \dfrac{3}{2}} \right\}$
Now displacement is the rate of change of velocity with respect to time. So it can be written that
\[\Rightarrow s = \int_0^5 {vdt} \]
$ \Rightarrow s = \int\limits_0^{\dfrac{3}{2}} {vdt} + \int\limits_{\dfrac{3}{2}}^5 {vdt} $
$ \Rightarrow \int\limits_0^{\dfrac{3}{2}} {( - 2t + 3)dt + \int\limits_{\dfrac{3}{2}}^5 {(2t - 3)dt} } $
$ \Rightarrow [\dfrac{{ - 2{t^2}}}{2} + 3t]_0^{\dfrac{3}{2}} + [\dfrac{{2{t^2}}}{2} - 3t]_{\dfrac{3}{2}}^5$
Substituting the lower and upper limits, and solving we get
$ \Rightarrow [ - \dfrac{9}{4} + \dfrac{9}{2}] + [25 - 15 - \dfrac{9}{4} + \dfrac{9}{2}]$
$ \Rightarrow - \dfrac{9}{2} + \dfrac{9}{2} + 10 + \dfrac{9}{2}$
$ \Rightarrow 10 + \dfrac{9}{2} = \dfrac{{20 + 9}}{2}$
$ \Rightarrow \dfrac{{29}}{2}$
Therefore, the displacement of the particle in a time period of $5s$ is $ \Rightarrow 14.5m$.
Note: It is important to note that the displacement of the particle has both magnitude and direction. So it is a vector quantity. Also the terms distance and displacement are not to be confused. They are both different terms. The displacement of the particle does not depend on the path and does not give any information about the path followed but the distance changes according to the path taken by the particle. Also the distance travelled by a particle can never be zero or negative. It is always positive. But the displacement of a particle can be positive, zero or negative.
Complete step by step solution:
Given that the speed depends as $v = \left| {2t - 3} \right|$
Therefore, it can be written that
$\Rightarrow v = \left\{ {2t - 3,t \geqslant \dfrac{3}{2}} \right\}$
Or $v = \left\{ { - 2t + 3,t < \dfrac{3}{2}} \right\}$
Now displacement is the rate of change of velocity with respect to time. So it can be written that
\[\Rightarrow s = \int_0^5 {vdt} \]
$ \Rightarrow s = \int\limits_0^{\dfrac{3}{2}} {vdt} + \int\limits_{\dfrac{3}{2}}^5 {vdt} $
$ \Rightarrow \int\limits_0^{\dfrac{3}{2}} {( - 2t + 3)dt + \int\limits_{\dfrac{3}{2}}^5 {(2t - 3)dt} } $
$ \Rightarrow [\dfrac{{ - 2{t^2}}}{2} + 3t]_0^{\dfrac{3}{2}} + [\dfrac{{2{t^2}}}{2} - 3t]_{\dfrac{3}{2}}^5$
Substituting the lower and upper limits, and solving we get
$ \Rightarrow [ - \dfrac{9}{4} + \dfrac{9}{2}] + [25 - 15 - \dfrac{9}{4} + \dfrac{9}{2}]$
$ \Rightarrow - \dfrac{9}{2} + \dfrac{9}{2} + 10 + \dfrac{9}{2}$
$ \Rightarrow 10 + \dfrac{9}{2} = \dfrac{{20 + 9}}{2}$
$ \Rightarrow \dfrac{{29}}{2}$
Therefore, the displacement of the particle in a time period of $5s$ is $ \Rightarrow 14.5m$.
Note: It is important to note that the displacement of the particle has both magnitude and direction. So it is a vector quantity. Also the terms distance and displacement are not to be confused. They are both different terms. The displacement of the particle does not depend on the path and does not give any information about the path followed but the distance changes according to the path taken by the particle. Also the distance travelled by a particle can never be zero or negative. It is always positive. But the displacement of a particle can be positive, zero or negative.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

Alpha, Beta, and Gamma Decay Explained

Alpha Particle Scattering and Rutherford Model Explained

Angular Momentum of a Rotating Body: Definition & Formula

Apparent Frequency Explained: Formula, Uses & Examples

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

