If a body travels equal displacements in equal intervals of time, then the velocity of body is uniform velocity. On the other hand, if the body covers unequal displacements in equal intervals of time, then its velocity is variable velocity.

A particle moving along a straight line or a path is said to undergo one dimensional motion.

Both represent negative velocity.

Speed is distance covered per unit time. Since distance cannot be negative therefore speed cannot be negative.

Slope of x-t graph provides velocity of motion. The nature of motion is identified by the shape of the graph.

An object is said to be point object if its dimensions are negligible as compared to the distance travelled by it. For example, an aeroplane which flies from Delhi to London.

Average speed=\(\frac { Total\ distance }{ Time\ taken } \)
Let x be the distance to be covered
∴ average speed = \(\frac { x }{ \frac { x }{ 2v_{ 1 } } +\frac { x }{ 2v_{ 2 } } } \)
where, \(\frac { x }{ 2v_{ 1 } } \) = time taken to cover first half of the distance
\(\frac { x }{ 2v_{ 2 } } \) = time taken to cover second half of the distance
Now, average speed =\(\frac { x }{ \frac { X(V_{ I }+V_{ 2 }) }{ 2V_{ I }V_{ 2 } } } \)
=\(\frac { x\times 2v_{ 1 }v_{ 2 } }{ x(v_{ 1 }+v_{ 2 }) } =\frac { 2v_{ 1 }v_{ 2 } }{ v_{ 1 }v_{ 2 } } \)
\(\Rightarrow \) V_av=\(\frac { 2\times 20m/s\times 60m/s }{ 100m/s } \)
= 48ms^-1

Here, distance covered by the car = (2500-1700)km
= 800 km
Time elapsed = 16 h
Average speed \(=\frac { Total \ distance }{ Toatal \ time } \)
\(u=\frac { 800 }{ 16 } =50\quad km/h\)
As 1 km = 1000 m and 1 h = 3600 s
\(u=\frac { 50\times 1000 }{ 1\times 3600 } =13.9\quad ms^{ -1 }\)

Let x bet the distance travelled in time t. Then , 
 x \(\propto\) t²
x = kt² [ here, k = constant of proportionality ]
We know that velocity  is gigven 
v = \( \frac{dx}{dt}\)
= 2kt
and acceleration is given by
a = \(\frac{dv}{dt}\) = 2 k  [ constant ]
thus, the body has uniform accelerated motion.

As given, x = 2t² + 3t
by differentiating  x w.r.t, we get 
Velocity,    \(v=\frac { dx }{ dt } =\frac { d }{ dt } ({ 2 }t^{ 2 }+3t)\)
v = (4t +3)
As velocity is time dependent, it means that motion is non-uniform

(i)  The velocity of a body at any instant.
(ii) The acceleration  of the body and 
(iii) the net displacement of the body in a given time - interval. 

The value of position x will be maximum, when the value of sin (wt) is maximum for this
\(sin(\omega t)=1=sin\quad \pi /2\)
or   \(\omega t=\frac { \pi }{ 2 } \Rightarrow t=\left( \frac { \pi }{ 2\omega } \right) \)

Let x be the displacement . Then,  x = at²
\(\therefore\)      Velocity of the object , v = \(\frac{dx}{dt}\)
Accleration  of the object , a = \(\frac{dv}{dt}\)
It means that a is constant.

An object could be considered as a point object if it covers a distance much larger than its own size.
e.g., If a bus of 5 m in size move 100 km, then the bus can be considered as a point object.