Question

The equation of motion of a particle is s = t3 − 3t, where s is in meters and t is in seconds. (Assume t ≥ 0.) (a) Find the velocity and acceleration as functions of t. g

Answer 1

Explanation:

The  equation of motion of a particle is :

$s=t^3-3t$

Where

s is in meters

t is in seconds

The position of a particle is, $v=\dfrac{ds}{dt}$

$v=(3t^2-3)\ m/s$

The acceleration of a particle is, $a=\dfrac{dv}{dt}$

$a=(6t)\ m/s^2$

Hence, this is the required solution.

Answer 2

Answer:

Velocity: $v(t) = 3t^{2} - 3$ m/s

Accelaration: $a(t) = 6t$ m/s²

Explanation:

The velocity is the derivative of the position.

The accelaration is the derivative of the velocity.

Derivative concepts

The derivative of $t^{n}$ is $n*t^{n-1}$

The derivative of a subtraction is the subtraction of the derivatives.

The derivative of a constant is 0.

The equation of the position is

$s(t) = t^{3} - 3t$ m

The equation of the velocity is

Derivative of the position

$v(t) = 3t^{2} - 3$ m/s

The equation of the acceleration is

$a(t) = 6t$ m/s²