The motion of an object along the $x$ -axis, the velocity $v$ depends on the displacement $x$ as $v = 3{x^2} - 2x$, then what is the acceleration at $x = 2m$.
A.$48m{s^{ - 2}}$
B.$80m{s^{ - 2}}$
C.$18m{s^{ - 2}}$
D.$10m{s^{ - 2}}$
Answer
535.2k+ views
Hint: The calculation of the shortest distance between the starting and the final point is called Displacement. It prefers a curved path or straight line. If the object moves in a straight path, the displacement is the result of the two different directions.
Formula used:
$ \Rightarrow a = v\dfrac{{dv}}{{dx}}$
Where $a$ is the displacement and $v$ is the velocity.
Complete answer:
Given, an object’s motion is along the $x$ -axis. The velocity of the object is given as $3{x^2} - 2x$ that is dependent on the displacement $x$. From the given information it is clear that the object is traveling in a straight line. To calculate the acceleration at $x = 2m$.
Acceleration is found by dividing the velocity by the displacement. That is,
$ \Rightarrow a = \dfrac{{dv}}{{dx}}$
The value of the velocity is given.
$ \Rightarrow a = v \times \dfrac{{d\left( {3{x^2} - 2x} \right)}}{{dx}}$
Differentiate the velocity value. That is,
$ \Rightarrow a = \left( {3{x^2} - 2x} \right)\dfrac{{d\left( {3{x^2} - 2x} \right)}}{{dx}}$
Differentiate the velocity value with respect to x we get,
$ \Rightarrow a = \left( {3{x^2} - 2x} \right)\left( {6x - 2} \right)$
The value of $x$ is given as $2m$. Substitute the value,
$ \Rightarrow a = \left( {3{{\left( 2 \right)}^2} - 2\left( 2 \right)} \right)\left( {6\left( 2 \right) - 2} \right)$
Simplify the terms.
\[ \Rightarrow a = \left( {12 - 4} \right)\left( {12 - 2} \right)\]
On further simplifications,
Multiply the values,
\[ \Rightarrow a = \left( 8 \right)\left( {10} \right)\]
\[ \Rightarrow a = 80m{s^{ - 2}}\]
Therefore, the acceleration at $x = 2m$ is $80m{s^{ - 2}}$.
Hence option $\left( B \right)$ is the correct answer.
Note:
There will always be confusion between the displacement and the distance. They seem to be the same but both have a different meaning. Displacement is the measure of how far the object is out. It defines the object’s change of position and it is a vector quantity. Whereas the distance is the object’s total movement and it is regardless of the direction of the object. Acceleration is with respect to time, it is the rate of change of velocity. It is also a vector quantity that has both magnitude and direction.
Formula used:
$ \Rightarrow a = v\dfrac{{dv}}{{dx}}$
Where $a$ is the displacement and $v$ is the velocity.
Complete answer:
Given, an object’s motion is along the $x$ -axis. The velocity of the object is given as $3{x^2} - 2x$ that is dependent on the displacement $x$. From the given information it is clear that the object is traveling in a straight line. To calculate the acceleration at $x = 2m$.
Acceleration is found by dividing the velocity by the displacement. That is,
$ \Rightarrow a = \dfrac{{dv}}{{dx}}$
The value of the velocity is given.
$ \Rightarrow a = v \times \dfrac{{d\left( {3{x^2} - 2x} \right)}}{{dx}}$
Differentiate the velocity value. That is,
$ \Rightarrow a = \left( {3{x^2} - 2x} \right)\dfrac{{d\left( {3{x^2} - 2x} \right)}}{{dx}}$
Differentiate the velocity value with respect to x we get,
$ \Rightarrow a = \left( {3{x^2} - 2x} \right)\left( {6x - 2} \right)$
The value of $x$ is given as $2m$. Substitute the value,
$ \Rightarrow a = \left( {3{{\left( 2 \right)}^2} - 2\left( 2 \right)} \right)\left( {6\left( 2 \right) - 2} \right)$
Simplify the terms.
\[ \Rightarrow a = \left( {12 - 4} \right)\left( {12 - 2} \right)\]
On further simplifications,
Multiply the values,
\[ \Rightarrow a = \left( 8 \right)\left( {10} \right)\]
\[ \Rightarrow a = 80m{s^{ - 2}}\]
Therefore, the acceleration at $x = 2m$ is $80m{s^{ - 2}}$.
Hence option $\left( B \right)$ is the correct answer.
Note:
There will always be confusion between the displacement and the distance. They seem to be the same but both have a different meaning. Displacement is the measure of how far the object is out. It defines the object’s change of position and it is a vector quantity. Whereas the distance is the object’s total movement and it is regardless of the direction of the object. Acceleration is with respect to time, it is the rate of change of velocity. It is also a vector quantity that has both magnitude and direction.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

10 examples of friction in our daily life

