Acceleration

The velocity of a vehicle on the road is never uniform. Its velocity increases and decreases at random. As the velocity of the vehicle on the road keeps on changing, it is said to have accelerated motion.

The time rate of change of velocity of an object is called the acceleration of the object. Thus, acceleration is the change in velocity in each unit of time. In case the change in velocity in each unit of time is constant, the object is said to be moving with constant acceleration, and such motion is called uniformly accelerated motion. On the other hand, if the change in velocity in each unit of time is not constant, the object is said to be moving with variable acceleration. Such a motion is called a non-uniformly accelerated motion. In general, the motion of a vehicle on the road is non-uniformly accelerated as it speeds up and slows down at random.

Acceleration

Since, the acceleration of an object is the rate of change of velocity of an object with respect to time, it may also be written mathematically as

$ \displaystyle \text{acceleration }=\dfrac{{\text{change of velocity}}}{{\text{time taken}}}$

Acceleration is a vector quantity. It is positive if the velocity is increasing and is negative if the velocity is decreasing. The negative acceleration is also called retardation or deceleration.

The unit of acceleration is $ \displaystyle \text{m/}{{\text{s}}^{2}}$ in MKS system or S.I and $ \displaystyle \text{cm/}{{\text{s}}^{2}}$ in CGS system. The dimensional formula of acceleration is $ \displaystyle [{{M}^{0}}{{L}^{1}}{{T}^{{-2}}}]$.

Uniform acceleration

An object is said to be moving with a uniform acceleration if its velocity changes by equal amounts in equal intervals of time.

Formulae for uniformly accelerated motion along a straight line

Consider an object, moving with a uniform acceleration ‘a‘ along a straight line OX, with origin at O. Let the object reach at points A and B at instants $ \displaystyle {{t}_{1}}\text{ and }{{t}_{2}}$. Let $ \displaystyle {{x}_{1}}\text{ and }{{x}_{2}}$ be the displacements of the object at time $  \displaystyle {{t}_{1}}\text{ and }{{t}_{2}}$ respectively and $ \displaystyle {{v}_{1}}\text{ and }{{v}_{2}}$ be the velocity of the object at position A and B respectively.

(i) Velocity-time relation

$  \displaystyle \because \text{acceleration }=\dfrac{{\text{change of velocity}}}{{\text{time taken}}}$

$ \displaystyle \begin{array}{l}\therefore a=\dfrac{{{{v}_{2}}-{{v}_{1}}}}{{{{t}_{2}}-{{t}_{1}}}}\text{ }…\text{(1)}\\\Rightarrow {{v}_{2}}-{{v}_{1}}=a\left( {{{t}_{2}}-{{t}_{1}}} \right)\\\Rightarrow {{v}_{2}}={{v}_{1}}+a\left( {{{t}_{2}}-{{t}_{1}}} \right)\text{ }…\text{(2)}\end{array}$

Let origin of time axis be taken at A and $ \displaystyle u$ be the velocity of the object at A and $  \displaystyle v$ be velocity of the object at B after time t. Then

$ \displaystyle \begin{array}{l}{{v}_{1}}=u;{{t}_{1}}=0;{{v}_{2}}=v\\\text{and }{{t}_{2}}=t\end{array}$

Putting values in (2), we get

$  \displaystyle v=u+at$

The relations (2) and (3) are required velocity-time relations.

(ii) Position-time relation

Let $ \displaystyle {{v}_{{av}}}$ be the average velocity of the object during the motion of the object from A to B,

$ \displaystyle {{v}_{{av}}}=\dfrac{{\text{total displacement}}}{{\text{total time taken}}}=\dfrac{{{{x}_{2}}-{{x}_{1}}}}{{{{t}_{2}}-{{t}_{1}}}}$

$ \displaystyle \begin{array}{l}\text{or }{{x}_{2}}-{{x}_{1}}={{v}_{{av}}}\left( {{{t}_{2}}-{{t}_{1}}} \right)\\\text{But, }{{v}_{{av}}}=\dfrac{{\text{initial velocity + final velocity}}}{2}\\=\dfrac{{{{v}_{1}}+{{v}_{2}}}}{2}\end{array}$

$ \displaystyle {\therefore {{x}_{2}}-{{x}_{1}}=\left( {\dfrac{{{{v}_{1}}+{{v}_{2}}}}{2}} \right)\left( {{{t}_{2}}-{{t}_{1}}} \right)\text{ }…\text{(4)}}$

$ \displaystyle {\text{or }{{x}_{2}}={{x}_{1}}+\left( {\dfrac{{{{v}_{1}}+{{v}_{2}}}}{2}} \right)\left( {{{t}_{2}}-{{t}_{1}}} \right)}$

$ \displaystyle {\because {{v}_{2}}={{v}_{1}}+a\left( {{{t}_{2}}-{{t}_{1}}} \right)}$

$ \displaystyle \begin{array}{l}\therefore {{x}_{2}}={{x}_{1}}+\\\left( {\dfrac{{{{v}_{1}}+\left( {{{v}_{1}}+a\left( {{{t}_{2}}-{{t}_{1}}} \right)} \right)}}{2}} \right)\left( {{{t}_{2}}-{{t}_{1}}} \right)\end{array}$

$ \displaystyle \begin{array}{l}\Rightarrow {{x}_{2}}={{x}_{1}}+\\{{v}_{1}}\left( {{{t}_{2}}-{{t}_{1}}} \right)+\frac{1}{2}a{{\left( {{{t}_{2}}-{{t}_{1}}} \right)}^{2}}\ldots \text{(5)}\end{array}$

If $ \displaystyle {{x}_{0}},u$ are the displacement and velocity of the object at $ \displaystyle {{t}_{1}}=0$ and x be the displacement of object at t, then using $ \displaystyle {{t}_{1}}=0,\text{ }{{x}_{1}}={{x}_{0}},\text{ }{{v}_{1}}=u,\text{ }{{t}_{2}}=t\text{ and }{{x}_{2}}=x$ in relation (5), we have

$ \displaystyle \begin{array}{l}x={{x}_{0}}+u\left( {t-0} \right)+\dfrac{1}{2}a{{\left( {t-0} \right)}^{2}}\\\Rightarrow x={{x}_{0}}+ut+\dfrac{1}{2}a{{t}^{2}}\text{ }…\text{(6)}\\\Rightarrow x-{{x}_{0}}=ut+\dfrac{1}{2}a{{t}^{2}}\text{ }…\text{(7)}\end{array}$

Using, $ \displaystyle x-{{x}_{0}}=S=$ distance travelled in time t, we have

$ \displaystyle S=ut+\dfrac{1}{2}a{{t}^{2}}\text{ }…\text{(8)}$

The relation (5), (6) and (8) are required position-time relations.

Using relation (6), the position-tie graph for positive and negative accelerations are shown in the image below. These graphs have parabolic shapes.

Positive acceleration:

Negative acceleration (Deceleration):

(iii) Position-velocity relation

Since,

$ \displaystyle {{v}_{2}}={{v}_{1}}+a\left( {{{t}_{2}}-{{t}_{1}}} \right)$

we have

$ \displaystyle \left( {{{t}_{2}}-{{t}_{1}}} \right)=\dfrac{{{{v}_{2}}-{{v}_{1}}}}{a}$

Putting this value in equation (4), we get

$ \displaystyle \begin{array}{*{20}{l}} \begin{array}{l}{{x}_{2}}-{{x}_{1}}=\left( {\frac{{{{v}_{1}}+{{v}_{2}}}}{2}} \right)\left( {\frac{{{{v}_{2}}-{{v}_{1}}}}{a}} \right)\\\Rightarrow {{x}_{2}}-{{x}_{1}}=\frac{{v_{2}^{2}-v_{1}^{2}}}{{2a}}\end{array} \\ {\Rightarrow v_{2}^{2}-v_{1}^{2}=2a({{x}_{2}}-{{x}_{1}})\text{ }…\text{(9)}} \end{array}$

If u and v are the velocities of an object at positions $ \displaystyle {{x}_{0}}\text{ and }x$ respectively, then using $ \displaystyle {{v}_{1}}=u$,

$ \displaystyle {{v}_{2}}=v,\text{ }{{x}_{1}}={{x}_{0}}\text{ and }{{x}_{2}}=x$ in (9), we get

$ \displaystyle {{v}^{2}}-{{u}^{2}}=2a\left( {{{x}_{0}}-x} \right)\text{ }…\text{(10)}$

If $ \displaystyle x-{{x}_{0}}=S$, then

$ \displaystyle {{v}^{2}}={{u}^{2}}+2aS\text{ }…\text{(11)}$

The relations (9), (10) and (11) are the required position-velocity relations.

Variable acceleration

An object is said to be moving with a variable acceleration if its velocity changes by unequal amounts in equal intervals of time.

Average acceleration

When an object is moving with a variable acceleration, then the average acceleration of the object for the given motion is defined as the ratio of the total change in velocity of the object during motion to the total time taken i.e.,

Average acceleration $ \displaystyle =\dfrac{{\text{total change of velocity}}}{{\text{total time taken}}}$

Instantaneous acceleration

When an object is moving with variable acceleration, then the object possesses different accelerations at different instants. The acceleration of the object at a given instant of time or at a given point of motion, is called its instantaneous acceleration.

Consider an object moving with non-uniform acceleration along a straight line. Let $ \displaystyle \overrightarrow{v}$ and $ \displaystyle \overrightarrow{v}+\Delta \overrightarrow{v}$ be the velocities of the object at instants of time $ \displaystyle t$ and $ \displaystyle t+\Delta t$, there $ \displaystyle \Delta t$ is very small time interval. The acceleration of the object at an instant $ \displaystyle t$ i.e., instantaneous acceleration is nearly equal to average acceleration in small time interval $ \displaystyle \Delta t$, because in this time-interval the acceleration of the object can be considered to be uniform, even though the object has non-uniform acceleration.

Therefore, instantaneous acceleration ‘a’ will be approximately equal to average acceleration in small time $ \displaystyle t+\Delta t$ and in given by

$ \displaystyle \overrightarrow{a}=\dfrac{{\Delta \overrightarrow{v}}}{{\Delta t}}$

The sense of approximation in the above expression can be removed by making the time interval $ \displaystyle \Delta t$ tends to zero. Thus, instantaneous acceleration

$ \displaystyle \overrightarrow{a}=\underset{{\Delta t\to 0}}{\mathop{{\lim }}}\,\left( {\dfrac{{\Delta \overrightarrow{v}}}{{\Delta t}}} \right)$

Hence, instantaneous acceleration of an object at a given instant is defined as the limiting value of the average acceleration in a small time interval around the given instant, when the time interval tends to zero.

Therefore, instantaneous acceleration,

$ \displaystyle \overrightarrow{a}=\underset{{\Delta t\to 0}}{\mathop{{\lim }}}\,\left( {\dfrac{{\Delta \overrightarrow{v}}}{{\Delta t}}} \right)=\dfrac{{dv}}{{dt}}$

$ \displaystyle \begin{array}{l}\text{As, }\overrightarrow{v}=\dfrac{{d\overrightarrow{x}}}{{dt}},\text{ therefore,}\\\overrightarrow{a}=\dfrac{d}{{dt}}\left( {\dfrac{{d\overrightarrow{x}}}{{dt}}} \right)=\dfrac{{{{d}^{2}}x}}{{d{{t}^{2}}}}\end{array}$

Thus, instantaneous acceleration of an object is also equal to the second derivative of the position of the object at the given instant.