** **

** 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:

\[v(t)=\int a(t) dt\]

**Expert Answer**

**Velocity and acceleratio****n** 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 acceleratio**n 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:

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