Question

Displacement is given by x = 2+ 3t2 + 5t3

. Find the value of instantaneous acceleration.​

Answer 1

AnswEr :

• Displacement of a particle is given by the equation, x = (5t³ + 3t² + t) m

We've to calculate instantaneous acceleration of the particle.

For calculating instantaneous acceleration we need to first find instantaneous velocity, by differentiating this equation, after that on again differentiating velocity we can find value of instantaneous acceleration. Or simply by finding double derivative we can get instantaneous acceleration.

Let's move ahead to question,

$\longrightarrow \tt{x\ =\ 5t^3\ +\ 3t^2\ +\ 2}$

Differentiate x with respect to time (t)

$\\ \longrightarrow \tt{\dfrac{dx}{dt}\ =\ \dfrac{d(5t^3\ +\ 3t^2\ +\ 2)}{dt}} \\ \\ \\ \longrightarrow \tt{v\ =\ 3(5)t^2\ +\ 2(3)t\ +\ 0 \; \; \; \; \; \; \; \; \; \bigg[\because\ v\ =\ \dfrac{dx}{dt} \bigg]} \\ \\ \\ \longrightarrow \underline{\boxed{\frak{v\ =\ (15t^2\ +\ 6t)\ ms^{-1}}}}$

Now, again differentiating the above equation with time ,

$\\ \longrightarrow \tt{\dfrac{dv}{dt}\ =\ \dfrac{d(15t^2\ +\ 6t)}{dt}} \\ \\ \\ \longrightarrow \tt{a\ =\ 2(15)t\ +\ 6 \; \; \; \; \; \; \; \; \; \bigg[\because\ a\ =\ \dfrac{dv}{dt} \bigg]} \\ \\ \\ \longrightarrow \underline{\boxed{\frak{a\ =\ (30t\ +\ 6)\ ms^{-2}}}}$