Question

The equation of motion of a particle is s(t) = t

Answer 1

Answer:

The velocity function is $v(t) = 3t^2-27 \cfrac ms$ and the acceleration function is $a(t) = 6t \cfrac{m}{s^2}$ .

Acceleration after 8 seconds is 48 meters per squared second.

Acceleration when the velocity is 0 is 18 meters per squared second.

And velocity is 0 m/s when the position is at the minimum.

Step-by-step explanation:

In order to find the velocity and the acceleration functions given the equation of motion, we need to find the first and second derivatives.

a) Finding the velocity function.

The velocity equation is given by

$v(t) = \cfrac{ds(t)}{dt}$

So for the motion equation of the exercise $s(t) = t^3-27t$ we have

$v(t) = \cfrac{d}{dt} (t^3-27t)$

$\boxed{v(t) = 3t^2-27 \cfrac ms}$

Finding the acceleration function.

The acceleration function is given by

$a(t) = \cfrac{dv(t)}{dt}$

Using the previous result give us

$a(t) = \cfrac{d}{dt} (3t^2-27)$

$\boxed{a(t) = 6t \cfrac m{s^2}}$

b) Finding the acceleration after 8 seconds.

In order to find the acceleration, we can plug the time t = 8 seconds on the acceleration equation

$a(t) = 6t \\a(8) = 6(8)\\a(8)= 48 \cfrac{m}{s^2}$

So the acceleration after 8 seconds is 48 meters per squared second

c) Finding the acceleration when the velocity is 0.

We can set first the velocity equal to and solve for the time t.

$v(t) = 0\\3t^2-27=0$

Solving for t

$3t^2 =27\\t^2 =9\\t = \pm 3$

It is important to notice that we only take into consideration the positive value of t, since we have as assumption $t\ge 0$.

Replacing t = 3 seconds on the acceleration equation.

$a(t) = 6t \\a(3) = 6(3)\\a(3)= 18 \cfrac{m}{s^2}$

The acceleration is 18 meters per squared second when the velocity is 0 m/s

d) Finding velocity when position is minimal

The position is minimal when the first derivative is equal to 0, since the derivative represents the velocity.

Thus we know that the velocity is going to be 0 when the position is at the minimum.