Velocity and position of the particle as a function of time

[shahin93]

Homework Statement

A particle starts from the origin with velocity 5 m/s at t = 0 and moves in the xy plane with a varying acceleration given by a = (9*squareroot*t)j, where "a" is in meters per second squared and t is in seconds.

Determine the velocity of the particle as a function of time.
Determine the position of the particle as a function of time

Homework Equations

I have no idea

The Attempt at a Solution

I have no idea

 

[danielakkerma]

Hi!
So you first might like to recall how does one determine acceleration, velocity and all that:
Remember that:
$\large<br /> \frac{d^2\vec{r}}{dt^2} = \vec{a}$
Split the components;
The acceleration is given only in the $\hat{j}$ direction, so that you're left with the simple task of integrating
$\frac{dv_y}{dt} = 9\sqrt{t}$
Try it,
Daniel

 

[shahin93]

Hi!
So you first might like to recall how does one determine acceleration, velocity and all that:
Remember that:
$\large<br /> \frac{d^2\vec{r}}{dt^2} = \vec{a}$
Split the components;
The acceleration is given only in the $\hat{j}$ direction, so that you're left with the simple task of integrating
$\frac{dv_y}{dt} = 9\sqrt{t}$
Try it,
Daniel

i tried that, i didnt get the right answer, can u show me your steps?

 

[danielakkerma]

Certainly!
Look here:
The velocity vector has two components, i, and j; in the i direction, it was initially constant, and with the abscence of acceleration in that direction it will remain so.
In in the j direction, after integration, we get:
$\large<br /> v_j = \displaystyle \int 9\sqrt{t}dt = 9\frac{2}{3}t^{\frac{3}{2}}=6t^{\frac{3}{2}}$
$\large<br /> v_i = 5$
$\large<br /> v = \sqrt{{v_i}^2 + {v_j}^2} = \sqrt{25+36t^3}$
And in order to evaluate the position, all you need is a simple exercise in calculus for $r=\displaystyle \int vdt$, again, by integrating each fraction of the velocity separately, and r = sqrt(r_i^2+r_j^2) as before...
Daniel