2.Motion in Straight Line
easy

Explain the acceleration.

Option A
Option B
Option C
Option D

Solution

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 $(\Delta v)$.

The unit of acceleration is $\mathrm{ms}^{-2}$.

– From average acceleration we cannot know how the velocity of particle changes with time.

Taking $\lim _{\Delta t \rightarrow 0}$ in equation then we get instantaneous acceleration $a$ at time $t$.

$\therefore a=\lim _{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{d v}{d t}$

Now, $v=\frac{d x}{d t}$ 

$\therefore a=\frac{d v}{d t}=\frac{d}{d t}\left(\frac{d x}{d t}\right)$

$\therefore a=\frac{d^{2} x}{d t^{2}}=x$

In other words second derivative of position with respect to time is acceleration of a particle.

If $\frac{d v}{d t}$ is positive, acceleration is along the positive X-axis and if $\frac{d v}{d t}$ is negative the acceleration is along the negative X-axis.

Standard 11
Physics

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