Question

A particle moves along straight line such that at time its position from a fixed point on the line is x = 3t^2
- 2
The velocity of the particle when t = 2 is​

Answer 1

Answer:

by differenciating the position by time we get velocity

v = 6t - 2

hence v at t= 2

==> v = 6 x 2 -2

= 12 -2

= 10 m/s

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Answer 2

✴ Correct Question :

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

▪ 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.