Physics, 19.07.2019 03:20 AnaiyaKirksey8

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

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Answer from: Jasoncookies23

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 from: memos5502

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

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