Another particle moving in a straight line problem

[karush]

A particle, moving in a straight line, passes through a fixed point O
with a velocity of $8 m s^{-1}$
Its acceleration, $a\ m^{-2}$, t seconds after passing O is given by $a=12-6t$. Find
(i) the velocity of the particle when $t=2$,
(ii) the displacement of the partial from O when $t=2$.

I know that if $a=12-6t$.
Then $v=12t+3t^2+C$

 

[Prove It]

A particle, moving in a straight line, passes through a fixed point O with a velocity of $8 m s^{-1}$
In acceleration, $a$ in $m\,\mathbf{s}^{-2}$, t seconds after passing O is given by $a=12-6t$. Find
(i) the velocity of the particle when $t=2$,
(ii) the displacement of the partial from O when $t=2$.

I know that if $a=12-6t$.
Then $v=12t+3t^2+C$

It would really help if you posted the question properly. I have fixed what I think you meant...

Anyway, since you know that when the point passes through the origin, the velocity is 8, that should help you work out your integration constant.

 

[MarkFL]

A particle, moving in a straight line, passes through a fixed point O
with a velocity of $8 m s^{-1}$
Its acceleration, $a\ m^{-2}$, t seconds after passing O is given by $a=12-6t$. Find
(i) the velocity of the particle when $t=2$,
(ii) the displacement of the partial from O when $t=2$.

I know that if $a=12-6t$.
Then $v=12t+3t^2+C$

You have been given the IVP:

$$a(t)=\d{v}{t}=12-6t$$ where $$v_0=8\frac{\text{m}}{\text{s}}$$

Another approach, would be to use the initial and final conditions (boundaries) as the limits of integration:

$$\int_{v_0}^{v(t)}\,du=\int_{t_0}^{t}12-6w\,dw$$

Now, we can choose to set the time at which the particle passes point $O$ as time $t=0$ and write:

$$\int_{8}^{v(t)}\,du=\int_{0}^{t}12-6w\,dw$$

Integrating, we find:

$$v(t)-8=12t-3t^2$$

Or:

$$v(t)=-3t^2+12t+8$$

Can you put together the IVP to find the position (displacement) function $x(t)$?

 

[karush]

$$v\left(2\right)=20 \frac{m}{s}$$
Which is the answer to (i)

But how do we get displacement if we don't have $s=$

 

[MarkFL]

$$v\left(2\right)=20 \frac{m}{s}$$
Which is the answer to (i)

But how do we get displacement if we don't have $s=$

We could let the initial position be 0, that is:

$$x(0)=0$$

But we don't have to...we could simply say $x(0)=x_0$, and then the displacement would be:

$$\Delta x=x(t)-x_0$$

 

[karush]

Well then $s\left(t\right)=-{t}^{3}+6{t}^{2}+8t$
Since $C=0$
So $s\left(2\right)=32 $

 

[MarkFL]

What I would do is write:

$$v(t)=\d{x}{t}=-3t^2+12t+8$$

Use the boundaries as the limits, and replace the dummy variables:

$$\int_0^{x(t)}\,du=\int_0^t -3w^2+12w+8\,dw$$

Utilize the FTOC:

$$x(t)=-t^3+6t^2+8t$$

And so:

$$x(2)=-8+24+16=32$$

And this agrees nicely with your result. :)