Question

For motion of an object along the x-axis,the velocity v depends on the displacement x as v=3x^2-2x,then what is the acceleration at x=2m.

Answer 1

Answer:

It is given in the question that velocity is a function of displacement.

Using the basic definitions of kinematics, we may state acceleration to be the change in velocity per unit time.

Hence in terms of velocity as a function of position, the derivative of this function will give us the instantaneous slope or change in velocity at that position.

Using this basic knowledge we may proceed with our problem

a(x)=dvdt

a(x)=dvdx∗dxdt

a(x)=v.dvdx (1)

Problem Statement

v(x)=3x2−2x;

Obtaining derivative for the function given,

dvdx=2.(3x)−2(1)=6x−2;

Multiplying the derivative of the function with the velocity function,

a(x)=v.dvdx=(6x−2)(3x2−2x);

Multiplying the terms,

a(x)=v.dvdx=18x3−12x2−6x2+4x

Simplifying,

a(x)=v.dvdx=18x3−18x2+4x

We now have an equation for acceleration as a function of position.

To find out the value of acceleration at given position, we may substitute values of x,

Here,

x=2;

Putting values in equation:

a(x)=18(2)3−18(2)2+4(2);

a(x)=18(8)−18(4)+4(2);

a(x)=144−72+8;

a(x)=80;

Therefore, acceleration at x=2m is 80ms−1

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Answer 2

Answer:

The coordinates of moving particle at any time ‘t’ are given by The speed of the particle at time ‘t’ is given by