Help needed with velocity and acceleration question

[lastochka]

Hello,
I need help with this question. I do have the answers, but I am not sure what is the logic behind these answers.
Here is the exercise
The equation of the motion of the particle is s=${t}^{2}$-3t, where s is in meters and t is in seconds. Find:a) the velocity and acceleration as function of t.
Logically I guess they asked about derivative, so here is what I did on my own, I took first derivative
f$^{\prime}$(x)=3${t}^{2}$-3 which is velocity? so according to my book second derivative is acceleration
a(t)+v$^{\prime}$(t)=6t
The answers are correct (from the book), but I don't understand why first derivative is velocity and second derivative is acceleration?? What is the logic behind it?

b)acceleration after 2 seconds
a(2)=6(2)=12$\frac{m}{{s}^{2}}$
Why are the seconds squared?I am very confused...help please.
Thank you

 

[MarkFL]

I am assuming you meant to given the position function as:

\(\displaystyle s(t)=t^3-3t\)

Now, using the definition that velocity $v$ is the time rate of change of position, we find:

\(\displaystyle v(t)=\d{s}{t}=3t^2-3\)

And similarly using the definition that acceleration $a$ is the time rate of change of velocity, we find:

\(\displaystyle a=\d{v}{t}=6t\)

These are just the way that velocity and acceleration are defined: velocity measures how fast position is changing and acceleration measures how fast velocity is changing.

If position is given in meters, and time measured in seconds, then velocity will be in meters per second $\dfrac{\text{m}}{{s}}$ and acceleration will be in meters per second per second $\dfrac{\frac{\text{m}}{\text{s}}}{\text{s}}=\dfrac{\text{m}}{\text{s}^2}$.

 

[lastochka]

Thank you! It is so clear now :)