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Question: The motion of an object along the \(x\) -axis, the velocity \(v\) depends on the displacement \(x\) ...

The motion of an object along the xx -axis, the velocity vv depends on the displacement xx as v=3x2−2xv = 3{x^2} - 2x, then what is the acceleration at x=2mx = 2m.
A.48ms−248m{s^{ - 2}}
B.80ms−280m{s^{ - 2}}
C.18ms−218m{s^{ - 2}}
D.10ms−210m{s^{ - 2}}

Explanation

Solution

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:
⇒a=vdvdx\Rightarrow a = v\dfrac{{dv}}{{dx}}
Where aa is the displacement and vv is the velocity.

Complete answer:
Given, an object’s motion is along the xx -axis. The velocity of the object is given as 3x2−2x3{x^2} - 2x that is dependent on the displacement xx. From the given information it is clear that the object is traveling in a straight line. To calculate the acceleration at x=2mx = 2m.
Acceleration is found by dividing the velocity by the displacement. That is,
⇒a=dvdx\Rightarrow a = \dfrac{{dv}}{{dx}}
The value of the velocity is given.
⇒a=v×d(3x2−2x)dx\Rightarrow a = v \times \dfrac{{d\left( {3{x^2} - 2x} \right)}}{{dx}}
Differentiate the velocity value. That is,
⇒a=(3x2−2x)d(3x2−2x)dx\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,
⇒a=(3x2−2x)(6x−2)\Rightarrow a = \left( {3{x^2} - 2x} \right)\left( {6x - 2} \right)
The value of xx is given as 2m2m. Substitute the value,
⇒a=(3(2)2−2(2))(6(2)−2)\Rightarrow a = \left( {3{{\left( 2 \right)}^2} - 2\left( 2 \right)} \right)\left( {6\left( 2 \right) - 2} \right)
Simplify the terms.
⇒a=(12−4)(12−2)\Rightarrow a = \left( {12 - 4} \right)\left( {12 - 2} \right)
On further simplifications,
Multiply the values,
⇒a=(8)(10)\Rightarrow a = \left( 8 \right)\left( {10} \right)
⇒a=80ms−2\Rightarrow a = 80m{s^{ - 2}}
Therefore, the acceleration at x=2mx = 2m is 80ms−280m{s^{ - 2}}.

Hence option (B)\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.