Question

The velocity of a particle is moving on the x-axis is given by v=x^2-x where vin m/s and x in m. Find its acceleration in m/sec^2? when passing through the
point x = 2m​

Answer 1

Given :

▪ The velocity of a particle moving along x-axis is given by

$\bigstar\:\underline{\boxed{\bf{\bf{\gray{v=x^2-x}}}}}$

To Find :

▪ Acceleration of particle at x = 2.

Solution :

Instantaneous acceleration is given by

$\dashrightarrow\sf\:a=\dfrac{dv}{dt}\\ \\ \dashrightarrow\sf\:a=\dfrac{dv}{dt}\times \dfrac{dx}{dx}\\ \\ \dashrightarrow\sf\:a=\dfrac{dx}{dt}\times \dfrac{dv}{dx}\\ \\ \dashrightarrow\sf\:a=v\times \dfrac{dv}{dx}\\ \\ \dashrightarrow\sf\:a=(x^2-x)\times \dfrac{d(x^2-x)}{dx}\\ \\ \dashrightarrow\sf\:a=(x^2-x)\times (2x-1)\\ \\ \dag\bf\:\red{putting\:x=2,\:we\: get}\\ \\ \dashrightarrow\sf\:a=[(2)^2-2]\times [(2\times 2)-1]\\ \\ \dashrightarrow\sf\:a=(4-2)\times (4-1)\\ \\ \dashrightarrow\sf\:a=2\times 3\\ \\ \dashrightarrow\underline{\boxed{\bf{\blue{a=6\:ms^{-2}}}}}\:\orange{\bigstar}$