Define acceleration , average acceleration and instantaneous acceleration.
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 $t$.
Similar Questions
The velocity $u$ and displacement $r$ of a body are related as $u^2 = kr$, where $k$ is a constant. What will be the velocity after $1\, second$ ? (Given that the displacement is zero at $t = 0$)
The velocity $u$ and displacement $r$ of a body are related as $u^2 = kr$, where $k$ is a constant. What will be the velocity after $1\, second$ ? (Given that the displacement is zero at $t = 0$)
Equation of motion of a body is $\frac{d v}{d t}=-4 v+8$, where $v$ is the velocity in $m / s$ and $t$ is the time in second. Initial velocity of the particle was zero. Then,
Equation of motion of a body is $\frac{d v}{d t}=-4 v+8$, where $v$ is the velocity in $m / s$ and $t$ is the time in second. Initial velocity of the particle was zero. Then,
Draw the $x\to t$ graphs for positive, negative and zero acceleration.
Draw the $x\to t$ graphs for positive, negative and zero acceleration.