Question

The velocity v of a moving object varies with displacement X as v=x^2-2x,then the acceleration X=2m is

Answer 1

Answer:

80ms−1 is the answer of this question.

Explanation:

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

Using this basic knowledge we may proceed with our problem

a(x)=dvdt

a(x)=dvdx.dxdt

a(x)=v.dvdx (1)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