Average Velocity $=\frac{\text { displacement }}{\text { timeinterval }}$

For straight line motion, along $x$-axis, we have

$$ \mathrm{v}_{\mathrm{av}}=\overline{\mathrm{v}}=\langle\mathrm{v}\rangle=\frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}=\frac{\mathrm{x}_{\mathrm{f}}-\mathrm{x}_{\mathrm{i}}}{\mathrm{t}_{\mathrm{f}}-\mathrm{t}_{\mathrm{i}}} $$

The velocity at a particular instant of time is known as instantaneous velocity. The term "velocity" usually means instantaneous velocity.

$$ V_{\text {inst. }}=\lim \limits_{\Delta t \rightarrow 0}\left(\frac{\Delta x}{\Delta t}\right)=\frac{d x}{d t} $$

The position of a particle refers to its location in the space at a certain moment of time.

The change in the position of a moving object is known as displacement.

The change in position vector of a particle going from initial position (say $i$ ) to final position (say $f$ ) is called Displacement or Relative Position Vector of the particle. It is denoted by $\Delta \vec{r}$, such that

$$ \Delta \vec{r}=\vec{r}_f-\vec{r}_i $$

The length of the actual path travelled by a particle during a given time interval is called as distance.

$$ \text { Average Speed }=\frac{\text { distance travelled }}{\text { time interval }} $$

The dimension of velocity is $\left[\mathrm{LT}^{-1}\right]$ and its $\mathrm{SI}$ unit is $\mathrm{m} / \mathrm{s}$.

Average acceleration $=\frac{\text { change in velocity }}{\text { time interval }}$

Average acceleration is a vector quantity whose direction is same as that of the change in velocity.

$$ \overrightarrow{\mathrm{a}}_{\mathrm{av}}=\frac{\Delta \overrightarrow{\mathrm{v}}}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{v}}_{\mathrm{f}}-\overrightarrow{\mathrm{v}}_{\mathrm{i}}}{\Delta \mathrm{t}} $$

Since for a straight line motion the velocities are along a line, therefore

$$ \mathrm{a}_{\mathrm{av}}=\frac{\Delta \mathrm{v}}{\Delta \mathrm{t}}=\frac{\mathrm{v}_{\mathrm{f}}-\mathrm{v}_{\mathrm{i}}}{\mathrm{t}_{\mathrm{f}}-\mathrm{t}_{\mathrm{i}}} $$

The instantaneous acceleration of a particle is its acceleration at a particular instant of time.

Instantaneous Acceleration is

$$ \vec{a}_{\mathrm{ins}}=\vec{a}=\lim \limits_{\Delta t \rightarrow 0} \frac{\Delta \vec{v}}{\Delta t}=\frac{d \vec{v}}{d t} $$

So, $\vec{a}=\frac{d \vec{v}}{d t}=\frac{d^2 \vec{x}}{d t^2}$

Consider a particle moving in a straight line with constant acceleration $a$. If $u$ is the initial velocity, $v$ is the velocity at time $t, s$ is the displacement $(=\Delta x)$ in time $t$ and $s_n^{\text {th }}$ is the displacement in the $n^{\text {th }}$ second of motion, then equations governing the motion of such a particle are

$$ \begin{aligned} &v=u+a t \\\\ &s=\Delta x=u t+\frac{1}{2} a t^2 \\\\ &v^2-u^2=2 a s \\\\ &s=\frac{u+v}{2} t \\\\ &s_n^{\text {th }}=u+\frac{1}{2} a(2 n-1) \end{aligned} $$

(a) a - t graph : If acceleration a is constant, then a t- graph is a horizontal line as shown.

(b) v - t graph : For constant acceleration, v is a linear polynomial in terms of t . Hence v t- graph is a straight line of slope a as shown.

(c) x - t graph : For constant acceleration, x is a quadratic polynomial in terms of t . Hence x t- graph is a parabola.

(1) Zero Velocity : As position of particle is fixed at all the times, so the body is at rest.

Slope $=\frac{dx}{dt}=\tan \theta =\tan 0^{\circ }=0$

$\Rightarrow$ Velocity of particle is zero.



(2) Uniform Velocity : Here $\tan \theta$ is constant.

so, $\tan \theta =\frac{dx}{dt}=$ constant

$\Rightarrow$ Velocity of particle is constant.

(3) Non Uniform Velocity (Increasing with Time): In this case, as time is increasing, $\theta$ is also increasing.

$\Rightarrow \frac{dx}{dt}=\tan \theta \text{ also increases }$

Hence, velocity of particle is increasing.



(4) Non Uniform Velocity (Decreasing with Time) : In this case, as time increases, $\theta$ decreases.

$\Rightarrow \frac{dx}{dt}=\tan \theta$ also decreases.

Hence, velocity of particle is decreasing.

(1) Zero Acceleration : Velocity is constant

$\begin{array}{rl}& \Rightarrow \tan \theta =0\\ & \\ & \Rightarrow \frac{dv}{dt}=0\end{array}$

Hence, acceleration is zero i.e., $v-t$ graph is a straight line parallel to time axis.



(2) Uniform Acceleration : $\tan \theta$ is constant

$\Rightarrow \frac{dv}{dt}=$ constant

i.e., $v$ - $t$ graph is a straight line with positive slope.



(3) Uniform Retardation : Since $\theta >90^{\circ }$,

$\therefore$ $\tan \theta$ is constant and negative.

$\Rightarrow \frac{dv}{dt}=$ negative constant

i.e., $v-t$ graph is a straight line with negative slope.

(1) Constant Acceleration : $\tan \theta =0$

$\Rightarrow \frac{da}{dt}=0$

Hence acceleration is constant i.e., $a-t$ graph is a straight line parallel to time axis.



(2) Uniformly Increasing Acceleration : $\theta$ is constant.

$\begin{array}{rl}& 0^{\circ }<\theta <90^{\circ }\Rightarrow \tan \theta >0\\ & \\ \Rightarrow & \frac{da}{dt}=\tan \theta =\text{ constant }>0\end{array}$

Hence, acceleration is uniformly increasing with time i.e., $a-t$ graph is a straight line with positive slope.



(3) Uniformly Decreasing Acceleration : Since $\theta >90^{\circ }$

$\Rightarrow \tan \theta$ is constant and negative

$\Rightarrow \frac{da}{dt}=$ negative constant

Hence, acceleration is uniformly decreasing with time i.e., $a-t$ graph is a straight line with negative slope.

Slopes of v - t or s - t graphs can never be infinite at any point, because infinite slope of v - t graph means infinite acceleration. Similarly, infinite slope of s - t graph means infinite velocity. Hence, the following graphs are not possible :

At one time, two values of velocity or displacement are not possible. Hence, the following graphs are not right :

1.
Take initial direction of motion of the particle as positive.

Example :

(a) If a particle is dropped from a tower, then we can take downward direction (the direction of initial motion) as positive.

(b) If a particle is thrown vertically upwards from a tower, then we can take upward direction (the direction of initial motion) as positive.

2.

The quantities ( i.e. u , v , a and s ) along the initial direction of motion will be taken as positive whereas opposite to the initial direction of motion are taken as negative.

For example :

(a) For the particle dropped from the tower of height h, we have taken downward direction (the direction of initial motion) as positive. So, in this case we have

   u = 0, v = +v , a = +g and s = +h as all are in the downward direction.

(b) For the particle thrown vertically upwards from a tower of height h, we have taken upward direction (the direction of initial motion) as positive. So, in this case, since

• initial velocity u is upwards, so u = +u ,

• acceleration due to gravity is downwards, so a = -g

3.

Now you can apply the equations of all one dimensional motion of the particle.