# Acceleration  (Page 2/2)

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$\begin{array}{l}\mathbf{a}=\frac{\Delta \mathbf{v}}{\Delta t}\end{array}$

The ratio denotes average acceleration, when the measurement involves finite time interval; whereas the ratio denotes instantaneous acceleration for infinitesimally small time interval, $\Delta t\to 0$ .

We know that velocity itself has the dimension of length divided by time; the dimension of the acceleration, which is equal to the change in velocity divided by time, involves division of length by squared time and hence its dimensional formula is $L{T}^{-2}$ . The SI unit of acceleration is $\mathrm{meter/}{\mathrm{second}}^{2}$ i.e. $\mathrm{m/}{s}^{2}$ .

## Average acceleration

Average acceleration gives the overall acceleration over a finite interval of time. The magnitude of the average acceleration tells us the rapidity with which the velocity of the object changes in a given time interval.

$\begin{array}{l}{\mathbf{a}}_{\mathrm{avg}}=\frac{\Delta \mathbf{v}}{\Delta t}\end{array}$

The direction of acceleration is along the vector $\Delta \mathbf{v}={\mathbf{v}}_{2}-{\mathbf{v}}_{1}$ and not required to be in the direction of either of the velocities. If the initial velocity is zero, then $\Delta \mathbf{v}={\mathbf{v}}_{2}=\mathbf{v}$ and average acceleration is in the direction of final velocity.

Difference of two vectors ${\mathbf{v}}_{2}-{\mathbf{v}}_{1}$ can be drawn conveniently following certain convention (We can take a mental note of the procedure for future use). We draw a straight line, starting from the arrow tip (i.e. head) of the vector being subtracted ${\mathbf{v}}_{1}$ to the arrow tip (i.e. head) of the vector from which subtraction is to be made ${\mathbf{v}}_{2}$ . Then, from the triangle law of vector addition, $\begin{array}{l}{\mathbf{v}}_{1}+\Delta \mathbf{v}={\mathbf{v}}_{2}\\ ⇒\Delta \mathbf{v}={\mathbf{v}}_{2}-{\mathbf{v}}_{1}\end{array}$

## Instantaneous acceleration

Instantaneous acceleration, as the name suggests, is the acceleration at a given instant, which is obtained by evaluating the limit of the average acceleration as $\Delta t\to 0$ .

$\begin{array}{ll}\mathbf{a}=\underset{\Delta t\to 0}{\mathrm{lim}\phantom{\rule{10pt}{0ex}}}\frac{\mathbf{\Delta }\mathbf{v}}{\Delta t}=& \frac{đ\mathbf{v}}{đt}\end{array}$

As the point B approaches towards A, the limit of the ratio evaluates to a finite value. Note that the ratio evaluates not along the tangent to the curve as in the case of velocity, but along the direction shown by the red arrow. This is a significant result as it tells us that direction of acceleration is independent of the direction of velocity.

Instantaneous acceleration
Instantaneous acceleration is equal to the first derivative of velocity with respect to time.

It is evident that a body might undergo different phases of acceleration during the motion, depending on the external forces acting on the body. It means that accelerations in a given time interval may vary. As such, the average and the instantaneous accelerations need not be equal.

A general reference to the term acceleration ( a ) refers to the instantaneous acceleration – not average acceleration. The absolute value of acceleration gives the magnitude of acceleration :

$\begin{array}{l}|\mathbf{a}|=a\end{array}$

## Acceleration in terms of position vector

Velocity is defined as derivative of position vector :

$\begin{array}{l}\mathbf{v}=\frac{đ\mathbf{r}}{đt}\end{array}$

Combining this expression of velocity into the expression for acceleration, we obtain,

$\begin{array}{ll}\mathbf{a}=\frac{đ\mathbf{v}}{đt}& =\frac{{đ}^{2}\mathbf{r}}{đ{t}^{2}}\end{array}$

Acceleration
Acceleration of a point body is equal to the second derivative of position vector with respect to time.

Now, position vector is represented in terms of components as :

$\begin{array}{l}\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\end{array}$

Substituting in the expression of acceleration, we have :

$\begin{array}{l}\mathbf{a}=\frac{{đ}^{2}x}{đ{t}^{2}}\mathbf{i}+\frac{{đ}^{2}y}{đ{t}^{2}}\mathbf{j}+\frac{{đ}^{2}z}{đ{t}^{2}}\mathbf{k}\\ \mathbf{a}={a}_{x}\mathbf{i}+{a}_{y}\mathbf{j}+{a}_{z}\mathbf{k}\end{array}$

## Acceleration

Problem : The position of a particle, in meters, moving in space is described by following functions in time.

$\begin{array}{l}x=2{t}^{2}-2t+3;y=-4t\phantom{\rule{3pt}{0ex}}\mathrm{and}\phantom{\rule{3pt}{0ex}}z=5\end{array}$

Find accelerations of the particle at t = 1 and 4 seconds from the start of motion.

Solution : Here scalar components of accelerations in x,y and z directions are given as :

$\begin{array}{l}{a}_{x}=\frac{{đ}^{2}x}{đ{t}^{2}}=\frac{{đ}^{2}}{đ{t}^{2}}\left(2{t}^{2}-2t+3\right)=4\\ {a}_{y}=\frac{{đ}^{2}y}{đ{t}^{2}}=\frac{{đ}^{2}}{đ{t}^{2}}\left(-4t\right)=0\\ {a}_{z}=\frac{{đ}^{2}z}{đ{t}^{2}}=\frac{{đ}^{2}}{đ{t}^{2}}\left(5\right)=0\end{array}$

Thus, acceleration of the particle is :

$\begin{array}{l}\mathbf{a}=4\mathbf{i}\end{array}$

The acceleration of the particle is constant and is along x-direction. As acceleration is not a function of time, the accelerations at t = 1 and 4 seconds are same being equal to $4m/{s}^{2}$ .

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