It is a uniform circular motion. So the direction of velocity changes but not the magnitude. Therefore speed in considered constant. Again magnitude of acceleration does not change.

\(\vec{v}=2 \hat{l}+t^{2} \hat{j}-9 \vec{k}\)
\(\vec{a}=\frac{d \vec{v}}{d t}=2 t \hat{j}\)
\(\text { When } t=0.5 \mathrm{~s}\)
\(a=1 \mathrm{~ms}^{-2}\)

Velocity and acceleration are in the same direction: So speed increases.

(i) Graphically the slope of the position-time graph will give the velocity of the particle.
(ii) At the same time, if velocity-time graph is given, the distance and displacement are determined, by calculating the area under the curve.
Velocity is given by \(\frac { dx }{ dt } =v\)
(iii) Therefore, dx = vdt
By integrating both sides, \(\int _{ { x }_{ 1 } }^{ { x }_{ 2 } }{ dx=\int _{ { x }_{ 1 } }^{ { x }_{ 2 } }{ v\quad dt } } \)
Integration is equivalent to area under the given curve.
(iv) So the term \(\int _{ { t }_{ 1 } }^{ { t }_{ 2 } }{ vdt } \) represents the area under the curve v as a function of time.
(v) Since the left hand side of the integration represents the displacement travelled by the particle from time t₁ to t₂, the area under the velocity time graph will give the displacement of the particle.

Displacement in the velocity - time graph
(vi) If the area is negative, it means that displacement is negative, so the particle has travelled in the negative direction.  

(i) In terms of position vector, the displacement vector is given as follows. Consider a particle moving from a point P₁ having position vector \(\overrightarrow { { r }_{ 1 } } ={ x }_{ 1 }\hat { i } +{ y }_{ 1 }\hat { j } +{ z }_{ 1 }\hat { k } \) to a point P₂ where its position vector is \(\overrightarrow { { r }_{ 2 } } ={ x }_{ 2 }\hat { i } +{ y }_{ 2 }\hat { j } +{ z }_{ 2 }\hat { k } \)
(ii) The displacement vector is given by \(\Delta \overrightarrow { r } =\overrightarrow { { r }_{ 2 } } -\overrightarrow { { r }_{ 1 } } \)
= (x₂-x₁)\(\hat { i } \) + (y₂-y₁)\(\hat { j } \)+(z₂-z₁)\(\hat {k } \)
(iii) This displacement is also shown in

360^o = 27 days
1 day = \(\frac{360^o}{27}=13^o.3'\)

360^o degrees in 24 hours.
\(\therefore\) angular displacement \(=\frac{360}{24}=15^{\circ}\ or \ \frac{2 \pi}{24}=\frac{\pi}{12}\) radius.

\(\overrightarrow { r } =3\hat { i } +2\hat { j } \)
\(\text { Unit vector } =\frac{\vec{r}}{|\vec{r}|}  \)
\(|\vec{r}| =\sqrt{3^{2}+2^{2}}=\sqrt{13}  \)
\(\therefore \text { Unit vector } =\frac{3 \hat{i}+2 \hat{j}}{\sqrt{13}}\)

If \(\theta \) in the angle made by the resultant acceleration, then \(tan \ \theta =\frac { { a }_{ 1 } }{ \left( \frac { { v }^{ 2 } }{ r } \right) } \) where a_t is the tangential acceleration and \(\frac{v^2}{r}\) is the centripetal acceleration.

1. \(\omega ={ \omega }_{ 0 }+\alpha t\)	\(\omega \) = Final angular velocity
2. \(\theta ={ \omega }_{ 0 }t+\frac { 1 }{ 2 } { \alpha t }^{ 2 }\)	\({ \omega }_{ 0 }\) = initial angular velocity
3. \({ \omega }^{ 2 }={ \omega }_{ 0 }^{ 2 }+2\alpha \theta \)	\(\theta \) = Angular displacement
4. \(\theta =\frac { \left( { \omega }_{ 0 }+\omega \right) t }{ 2 } \)	\(\alpha \) = angular acceleration t = time

Velocity:
Velocity is equal to the rate of change of position vector with respect to time. 
It is a vector quantity \(\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}\)
Speed:
The magnitude of velocity is called speed and is given by \(v= \sqrt{v^2_x+v^2_y+v^2_z}\). It is a positive scalar.

Position of a particle x = 3t² - 4t + 30
Velocity \(v=\frac { dx }{ dt } =\frac { d }{ dt } ({ 3t }^{ 2 }-4t+30)\)
= 6t - 4   [\(\because\) xⁿ = nx^n-1]
Acceleration \(a=\frac { dv }{ dt } =\frac { d }{ dt } (6t-4)\)
a = 6
At time t = 0, we have
Position x = 3t² - 4t + 30
=3(0)² - 4(0)+30
x = 30 m
Velocity v = 6t - 4
= 6(0) - 4
= -4 m/s = -4 ms^-1 
Acceleration a = 6 ms^-2
\(\because\) position x = 30 m, velocity v = -4 ms^-1, acceleration = 6 ms^-2

The position of a particle x = 6t + 2t³
By differentiating with respect to 't'
\(\frac { dx }{ dt } =\frac { d }{ dt } (6t+2{ t }^{ 3 })\)
\(\frac { dx }{ dt } =6+6{ t }^{ 2 }\)
\(\because \ velocity\ v=\frac { dx }{ dt } =6+6{ { t }^{ 2 } }\ \ [\because { x }^{ n }={ nx }^{ n-1 }]\)
N = 6 + 6 t²
As velocity is time independent, it means that motion is uniform

(i) Consider a particle moving along a circular path of radius r with a variable speed v.
(ii) As the speed of the particle changes so acceleration has a tangential component,
\({ a }_{ t }=\frac { dv }{ dt } r\infty ={ a }_{ t }={ r }_{ \infty }\)
(iii) As the direction of motion changes continuously, so the acceleration has a radial component (i.e.) centripetal acceleration.
\(\therefore \quad { a }_{ c }=\frac { { v }^{ 2 } }{ r } \)
(iv) The resultant acceleration is obtained by vector sum of centripetal and tangential acceleration.

(v) the magnitude of this resultant acceleration is given by \({ a }_{ R }=\sqrt { { a }_{ t }^{ 2 }+{ \left( \frac { { v }^{ 2 } }{ r } \right) }^{ 2 } } \)

Consider a uniform circular motion of a point object moving on a circular path of radius r. In uniform circular motion, the velocity vector turns continuously without changing its magnitude (speed). Let v be its velocity.
The velocity is tangential at every point in the circle. But the acceleration 'a' is acting towards the centre of the circle. This is called centripetal acceleration. It always points towards the centre of the circle.
Let the direction of position and velocity vectors shift through the same angle θ in a small interval of time \(\triangle t\) as shown in figure. For uniform circular motion \(r=|\overrightarrow{r_1}|={r_2}| \) and \(vr=|\overrightarrow{v_1}|=\overrightarrow{v_2}|\)

If the particle moves from position vector \(\overrightarrow{r_1}\) to \(\overrightarrow{r_2}\) , the displacement is \(\triangle{\overrightarrow{r}}={\overrightarrow{r}}_{2}-{\overrightarrow{r}}_{1}\) and change in velocity from \(\overrightarrow{v_1}\) to \(\overrightarrow{v_2}\) is \(\triangle={\overrightarrow{v}}_{1}={\overrightarrow{v}}_{2}-{\overrightarrow{v}}_{1}\) then the magnitude of the displacement \(\triangle r \ and \ of \triangle v\) satisfy the following relation:
\({{\triangle r}\over{r}}=-{{\triangle v}\over{v}}=\theta\)
Here the negative sign implies that \(\triangle r\) points radially inward, towards the centre of the circle
\(\triangle v=-v.{{\triangle r}\over{r}}\)
Then \(a={{\triangle r}\over{\triangle t}}={{r}\over{r}}\left( {{\triangle r}\over{\triangle t}} \right)=-{{v}^{2}{}\over{r}}\)
For uniform circular motion v = \(\omega\)r. where \(\omega\) - angular velocity of the object about the centre. Then the centripetal acceleration can be written as
\({a=-\omega^2 r}\)

(i) It is a vector which denotes the position of a particle at any instant of time, with respect to some reference frame or coordinate system.

(ii) The position vector \(\overrightarrow { r } \)of the particle at a point P is given by \(\overrightarrow { r } =x\overrightarrow { i } +y\overrightarrow { j } +z\overrightarrow { k } \) where x, y and z are components of \(\overrightarrow { r } \).

\(\overrightarrow { A } =4\hat { i } -2\hat { j } +\hat { k } \)
\(\overrightarrow { B } =5\hat { i } +3\hat { j } -4\hat { k } \)
Resultant vector = \(\overrightarrow { A } +\overrightarrow { B } \)
\(=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & -2 & 1 \\ 5 & 3 & -4 \end{matrix} \right| \)
\(=\hat{i}(8-3)+\hat{j}[5-(-16)]+\hat{k}(12+10)\)
Resultant vector = 5\(\hat { i } \) +21\(\hat { j } \) + 22\(\hat { k } \)