Question

A particle is moving along x-axis so that its velocity is given as v=(x^2+x+4) m/s, where x is in m. the acceleration of the particle at x=1.5m is

Answer 1

GiveN :

• Velocity of a particle is given as, $\sf{v\ =\ ( x^2\ +\ x\ +\ 4\ )\ ms^{-1}}$

To FinD :

• Acceleration of particle at x = 1.5 m

SolutioN :

Use Differentiation for calculating acceleration of particle :

$\implies \sf{a\ =\ v \dfrac{dv}{dx}} \\ \\ \\ \implies \sf{a\ =\ (x^2\ +\ x\ +\ 4) \dfrac{d(x^2\ +\ x\ +4)}{dx}} \\ \\ \\ \implies \sf{a\ =\ (x^2\ +\ x\ +\ 4) \: \times \: \big[ 2(x)\ +\ 1\ +\ 0 \big] } \\ \\ \\ \implies \sf{a\ =\ (x^2\ +\ x\ +\ 1) \: \times \: (2x\ +\ 1)} \\ \\ \\ \longrightarrow \underline{\boxed{\sf{Acceleration\ =\ (x^2\: +\ x\ +\ 1) \: \times \: (2x\ +\ 1)\ ms^{-2}}}}$

_______________________

Put x = 1.5 m

$\\ \implies \sf{a\ =\ \big[ (1.5)^2\ +\ 1.5\ +\ 4 \big] \: \times \: \big[ 2(1.5)\ +\ 1 \big]} \\ \\ \\ \implies \sf{a\ =\ (2.25\ +\ 1.5\ +\ 4) \: \times \: (3\ +\ 1)} \\ \\ \\ \implies \sf{a\ =\ 7.75 \: \times \: (4)} \\ \\ \\ \implies \sf{a\ =\ 11.75} \\ \\ \\ \longrightarrow \underline{\boxed{\sf{Acceleration\ (a)\ =\ 11.75 ms^{-2}}}}$