Question

A particle is moving along positive x - axis. Its position varies as a = t³ - 3t² + 12t+ 20, where x is in meters
and t is in seconds. Initial acceleration of the particle is
(B) 1 m/s2
(A) Zero
(C) -3 m/s2
(D)-6 m/s2​

Answer 1

Answer:

its answer is

Explanation:

A. ZERO

HOPE U UNDERSTAND

Answer 2

✴ Correct Question :

A particle moves along a straight line such that its displacement at any time t is given by $\bf{s=t^3-6t^2+3t+4} find the velocity when acceleration is zero.</p><p></p><h2>✴ Solution :</h2><p></p><p>✏ We know that,</p><p></p><p>[tex]\begin{lgathered}\bigstar\bf\:v=\dfrac{ds}{dt}\\ \\ \bigstar\bf\:a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}\end{lgathered}$

▪ Acceleration of particle :

$\begin{lgathered}\longrightarrow\sf\:a=\dfrac{d^2s}{dt^2}\\ \\ \longrightarrow\sf\:a=\dfrac{d^2(t^3-6t^2+3t+4)}{dt^2}\\ \\ \longrightarrow\sf\:a=6t-12\\ \\ \longrightarrow\sf\:ATQ,\:a=0\\ \\ \longrightarrow\sf\:a=6t-12=0\\ \\ \longrightarrow\sf\:6t=12\\ \\ \longrightarrow\boxed{\bf{\large{t=2\:s}}}\end{lgathered}$

▪ Velocity of particle at t = 2s

$\begin{lgathered}\longrightarrow\sf\:v=\dfrac{ds}{dt}\\ \\ \longrightarrow\sf\:v=\dfrac{d(t^3-6t^2+3t+4)}{dt}\\ \\ \longrightarrow\sf\:v=3t^2-12t+3\\ \\ \longrightarrow\sf\:v=3(2)^2-12(2)+3\\ \\ \longrightarrow\sf\:v=12-24+3\\ \\ \longrightarrow\boxed{\bf{\red{v=-9\:mps}}}\end{lgathered}$

▶ Additional information :

⏭ Velocity is a vector quantity.

⏭ It can be positive, negative and zero.