Define acceleration , average acceleration and instantaneous acceleration.
The time rate of change of velocity is called acceleration.
Let a particle be moving in a straight line and at time $t_{1}$ and $t_{2}$ its velocities are $v_{1}$ and $v_{2}$ respectively. Thus, the change in velocity of the particle in time interval $\Delta t=t_{2}-t_{1}$ is $v_{2}-v_{1}$. According to definition of average acceleration,
$\text { Average acceleration }=\frac{\text { change in velocity }}{\text { time }}$
$\therefore\langle a\rangle=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}=\frac{\Delta v}{\Delta t}$
Average acceleration is a vector quantity and its direction is in the direction of change in velocity $(\Delta v)$.
Taking $\lim _{\Delta t \rightarrow 0}$ in equation then we get instantaneous acceleration $a$ at time $t$.
A particle starts moving along a line from zero initial velocity and comes to rest after moving distance $d$. During its motion, it had a constant acceleration $f$ over $2 / 3$ of the distance and covered the rest of the distance with constant retardation. The time taken to cover the distance is
The relation between time ' $t$ ' and distance ' $x$ ' is $t=$ $\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. The relation between acceleration $(a)$ and velocity $(v)$ is:
If a body starts from rest and travels $120 \,cm$ in the $6^{th}$ second, then what is the acceleration.........$m/{s^2}$
Mark the correct statements for a particle going on a straight line
A particle starts moving with acceleration $2 \,m / s ^2$. Distance travelled by it in $5^{\text {th }}$ half second is ....... $m$