 # Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position.

a(t)= 2i+2kt, v(0)=3i-j, r(0)=j+k

This question aims to find a particle’s velocity and position vector with some acceleration, initial velocity, and position vectors. A position vector helps us find one object’s position relative to another. Position vectors normally start at the origin and end at any arbitrary point. Thus, these vectors are used to determine the position of a certain point relative to its source.

A position vector is a straight line with one end attached to a body, and the other attached to a moving point and is used to describe the position of a point relative to the body. As the point moves, the position vector will change in length, direction, or distance and direction. A position vector is a vector that shows either the position or location of any given point relative to any reference point, such as the origin. The direction of the position vector always points from the origin of this vector to the given point.

In a Cartesian coordinate system, if $O$ is the origin and $P(x1, y1)$ is the next point, then the position vector that is directed from $O$ to $P$ can be represented as $OP$.

In three-dimensional space, if the origin is $O = (0,0,0)$ and $P = (x_{1}, y_{1}, z_{1})$, then the position vector at $P$ can be represented as: $v = x_{1}i + y_{1}j + z_{1}k$.

Rate of change of the displacement is called velocity, while the rate of change of velocity is called acceleration.

The relationship between velocity and acceleration vector is:

$v(t)=\int a(t) dt$

Velocity and acceleration are related through the following formula:

$v(t)=\int a(t)dt$

The value of acceleration is given in the data.

$a(t)=2i+2kt$

Therefore,

$v(t)=\int 2i+2kt dt$

$v(t)=2it+kt^{2}+C$

Where $C$ represents the constant vector.

Given that:

$v(0)=3i-j$

$3i-j=C$

Plug value of $C$,

$v(t)=2it+kt^{2}+3i-j$

$v(t)=(2t+3)i-j+kt^{2}$

$r(t)=\int v(t)dt$

$r(t)=\int (2t+3)i-j+kt^{2} dt$

$r(t)=(t^{2}+3t)i-tj+k\dfrac{t^{3}}{3}+C$

$r(0)=j+k$

$r(t)=(t^{2}+3t)i-tj+k\dfrac{t^{3}}{3}+j+k$

The position vector is

$r(t)=(t^{2}+3t)i+(1-t)j+(\dfrac{t^{3}}{3}+1)k$

## Numerical Result

The velocity vector is given as:

$v(t)=(2t+3)i-j+kt^{2}$

The position vector is given as:

$r(t)=(t^{2}+3t)i+(1-t)j+(\dfrac{t^{3}}{3}+1)k$

## Example

Find the velocity and position vectors of a particle that has a given acceleration and a given initial velocity and position.

$a(t)=4i+4kt$, $v(0)=5i-j$, $r(0)=2j+k$

Solution

Velocity and acceleration are related through the following formula:

$v(t) = \int a(t)dt$

The value of acceleration is given in the data.

$a(t)=4i+4kt$

Therefore,

$v(t)=\int 4i+4kt dt$

$v(t)=4it+2kt^{2}+C$

Where $C$ represents the constant vector.

Given that:

$v(0)=5i-j$

$5i-j=C$

Plug value of $C$,

$v(t)=4it+2kt^{2}+5i-j$

$v(t)=(4t+5)i-j+2kt^{2}$

The position vector is:

$r(t)=(2t^{2}+5t)i+(2-t)j+(2\dfrac{t^{3}}{3}+1)k$

The velocity vector is given as:

$v(t)=(4t+5)i-j+2kt^{2}$

The position vector is given as:

$r(t)=(2t^{2}+5t)i+(2-t)j+(2\dfrac{t^{3}}{3}+1)k$