Question

Q17) A particle of unit mass undergoes one dimensional motion such that its velocity varies according to

v(x) = bx-2n, where b and n are constants and x is the position of the particle. Find acceleration as a function

of x.

Answer 1

A particle of unit mass undergoes one dimensional motion such that its velocity varies as $\bf{v(x)=bx^{-2n}}$

where b and n are constants and x is the position of the particle.

we have to find acceleration as a function of x.

we know, “acceleration is the rate of change of velocity with respect to time.”

i.e., a = dv/dt = dv/dx × dx/dt = dv/dx × v

so, a = v dv/dx

or, a(x) = $v(x)\frac{dv(x)}{dx}$

⇒a(x) = $bx^{-2n}\frac{d\left[bx^{-2n}\right]}{dx}$

⇒a(x) = $bx^{-2n}\times b(-2n)x^{-2n-1}$

⇒a(x) = $(-2nb^2)x^{-4n-1}$

hence, acceleration as a function of x, a(x) = $\bf{(-2nb^2)x^{(-4n-1)}}$

Answer 2

Explanation :

It is given that,

Velocity of particle, $v(x) = bx^{-2n}$

Where

b and n are constant

We know that,

$a=\dfrac{dv}{dt}$

or

$a=v\dfrac{dv}{dx}$...........(1)

$\dfrac{dv}{dx}=b(-2n)x^{-2n-1}$

So, equation (1) becomes :

$a=(bx^{-2n}).b$

or

$a=(-2n)b^2x^{-4n-1}}{$

Hence, this is the required solution.