Question

Let s(t)=6t^3−18t^2−54t be the equation of motion for a particle. Find a function for the velocity.
v(t)=

Where does the velocity equal zero? [Hint: factor out the GCF.]
t= and t=________

Find a function for the acceleration of the particle.
a(t)=___________

Answer 1

Answer:

Velocity: $v(t) = 18t^{2} - 36t - 54$

The velocity equals 0 at $t = 3$

Acceleration: $a(t) = 36t - 36$

Step-by-step explanation:

The velocity is the derivative of the position

The acceleration if the derivative of the velocity

The position is given by the following equation

$s(t) = 6t^{3} - 18t^{2} - 54t$

So the equation for the velocity is:

$v(t) = 18t^{2} - 36t - 54$

Where does the velocity equal zero? [Hint: factor out the GCF.]

This is when $v(t) = 0$.

So

$v(t) = 0$

$18t^{2} - 36t - 54 = 0$

The greatest common factor between 18,36 and 54 is 18. So, factoring by 18

$t^{2} - 2t - 3 = 0$

How we find the roots of a quadratic function?

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

So

$t^{2} - 2t - 3 = 0$

$a = 1, b = -2, c = -3$

$\bigtriangleup = b^{2} - 4ac = (-2)^{2} -4*1*(-3) = 16$

$t_{1} = \frac{-(-2) + \sqrt{16}}{2} = 3$

$t_{2} = \frac{-(-2) - \sqrt{16}}{2} = -1$

We do not have a negative instant of time, so the velocity equals 0 at $t = 3$

Find a function for the acceleration of the particle.

$v(t) = 18t^{2} - 36t - 54$

$a(t) = 36t - 36$

Answer 2

The function for the velocity is $v(t)=18t^2-36t-54$.

The velocity is zero at t = 3 and t = -1

The function for acceleration is $a(t)=36t-36$.

a) To find the function for velocity we'll find the first derivative of the distance function s(t).

$v(t)=\frac{d}{dt}s(t)\\ \\v(t)= \frac{d}{dt}(6t^3-18t^2-54t)\\\\v(t)=6(3t^2)-18(2t)-54(1)\\\\v(t)=18t^2-36t-54$

b) We set the velocity equal to 0 and solve for t.

$18t^2-36t-54=0\\18(t^2-2t-3)=0\\18(t-3)(t+1)=0$

We set each factor equal to 0 and solve for t.

$t-3=0\\t=3$

and

$t+1=0\\t=-1$

c) To find acceleration we differentiate v(t).

$a(t)=\frac{d}{dt} (18t^2-36t-54)\\a(t)=18(2t)-36(1)-0\\a(t)=36t-36$

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