(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) The instantaneous velocity at an instant or simply 'velocity' at an instant t is defined as limiting value of the average velocity as \(\Delta \)t \(\rightarrow\)0, evaluated at time t.
(ii) In other words, velocity is equal to rate of change of position vector with respect to time. Velocity is a vector quantity.
\(\overrightarrow { v } =\lim _{ \Delta x\rightarrow 0 }{ \frac { \Delta \overrightarrow { r } }{ \Delta \overrightarrow { t } } =\frac { d\overrightarrow { r } }{ dt } } \)
(iii) In component form, this velocity is
\(\overrightarrow { v } =\frac { d\overrightarrow { r } }{ dt } =\frac { d }{ dt } (x\hat { i } +y\hat { j } +z\hat { k } )\)
\(=\frac { dx }{ dt } \hat { i } =\frac { dy }{ dt } \hat { j } +\frac { dz }{ dt } \hat { k } \)
Here \(\frac { dx }{ dt } ={ v }_{ x }=x-\) component of velocity.
\(\frac { dx }{ dt } ={ v }_{ y }=y-\) component of velocity.
\(\frac { dz }{ dt } ={ v }_{ z }=z\) - component of velocity.
(iv) The magnitude of velocity v is called speed and is given by
\(v=\sqrt { { v }_{ x }^{ 2 }+{ v }_{ y }^{ 2 }+{ v }_{ z }^{ 2 } } \)
Speed is always a positive scalar. The unit of speed is also meter per second.

(i) The average velocity is defined as ratio of the displacement vector to the corresponding time interval
\(\overrightarrow { { v }_{ avg } } =\frac { \Delta \overrightarrow { r } }{ \Delta t } \)
(ii) It is a vector quantity. The direction of average velocity is in the direction of the displacement vector (\(\Delta \)\(\overrightarrow { r } \)).

(i) If a function is represented by y = f (x), then dy/dx represents the derivative of y with respect to x.
(ii) Mathematically this represents the variation of y with respect to change in x, for various continuous values of x.
(iii) Mathematically the derivative dy/dx is defined as follows
\(\frac { dy }{ dx } =\underset { \Delta x\rightarrow 0 }{ lim } \frac { y(x+\Delta x)-y(x) }{ \Delta x } \)
\(=\lim _{ \Delta x\rightarrow 0 }{ \frac { \Delta y }{ \Delta x } } \)
\(\frac { dy }{ dx } \) represents the limit that the quantity \(\frac { \Delta y }{ \Delta x } \)
attains, as \(\Delta \)x tends to zero. 

(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