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:
$\large \frac{d^2\vec{r}}{dt^2} = \vec{a}$
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:
$\large \frac{d^2\vec{r}}{dt^2} = \vec{a}$
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:
$\large v_j = \displaystyle \int 9\sqrt{t}dt = 9\frac{2}{3}t^{\frac{3}{2}}=6t^{\frac{3}{2}}$
$\large v_i = 5$
$\large v = \sqrt{{v_i}^2 + {v_j}^2} = \sqrt{25+36t^3}$
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

 

[asteriatic]

As a scientist, it is important to understand and apply the fundamental equations of motion to solve problems related to the position and velocity of a particle. In this case, we can use the equations of motion to determine the velocity and position of the particle as a function of time.

First, let's start with the equation for velocity: v = u + at. In this equation, v represents the final velocity, u represents the initial velocity, a represents the acceleration, and t represents time.

In this problem, we are given the initial velocity (u = 5 m/s) and the acceleration (a = 9*squareroot*t). Therefore, we can plug these values into the equation and get the velocity as a function of time: v = 5 + (9*squareroot*t).

Next, we can use the equation for position: s = ut + (1/2)at^2. In this equation, s represents the final position, u represents the initial velocity, a represents the acceleration, and t represents time.

Again, we are given the initial velocity (u = 5 m/s) and the acceleration (a = 9*squareroot*t). Therefore, we can plug these values into the equation and get the position as a function of time: s = 5t + (1/2)(9*squareroot*t)t^2.

By using these equations, we can determine the velocity and position of the particle at any given time. It is important to note that these equations assume that the acceleration is constant, which may not be the case for this problem. If the acceleration is not constant, we would need to use more advanced equations of motion to accurately calculate the velocity and position of the particle.