Velocity and position of the particle as a function of time

AI Thread Summary
The discussion focuses on determining the velocity and position of a particle moving in the xy plane with a given acceleration of a = (9√t)j. The initial velocity is 5 m/s at t = 0, and the acceleration only affects the j direction. After integrating the acceleration, the velocity in the j direction is found to be v_j = 6t^(3/2), while the velocity in the i direction remains constant at 5 m/s. The total velocity is calculated as v = √(25 + 36t^3). To find the position, integration of the velocity components is required, leading to the position function r = √(r_i^2 + r_j^2).
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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
 

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

i tried that, i didnt get the right answer, can u show me your steps?
 
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:
<br /> \large<br /> v_j = \displaystyle \int 9\sqrt{t}dt = 9\frac{2}{3}t^{\frac{3}{2}}=6t^{\frac{3}{2}} <br />
<br /> \large<br /> v_i = 5<br />
<br /> \large<br /> v = \sqrt{{v_i}^2 + {v_j}^2} = \sqrt{25+36t^3}<br />
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
 
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