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.