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Uniform Acceleration Linear Motion and its Graphs 📂Physics

Uniform Acceleration Linear Motion and its Graphs

Definition

When an object’s acceleration does not change over time tt, it is said to undergo uniform acceleration.

a(t)=a a(t)=a

Uniformly accelerated motion refers to moving in a straight line with constant, unchanging acceleration. What matters here is whether the acceleration aa is positive or negative: if a>0a>0, it will move faster in the initial direction, and if a<0a<0, it will slow down to a velocity of 0 and then start moving faster in the opposite direction. If a=0a=0, it means there is no change in velocity, indicating uniform motion.

Formulas

To reiterate, uniform acceleration implies constant acceleration. By integrating the acceleration to sequentially obtain velocity and position, below are three formulas derived. These formulas are frequently and usefully employed, so be sure to remember them.

v=adt=v0+atx=vdt=(v0+at)dt=x0+v0t+12at22ax=v2v02 \begin{align} v &= \int a dt = v_{0} + at \\ x &= \int v dt=\int(v_{0} + at)dt = x_{0} + v_{0}t + \frac{ 1 }{ 2 }at^{2} \\ 2ax &= v^{2}-v_{0}^{ 2 } \end{align}

Here, v0v_{0} and x0x_{0} are integration constants representing the initial values of velocity and position. (3)(3) is particularly used when there’s no given information about time tt in problems and can be derived from (1)(1) and (2)(2).

Derivation

First, rearranging (1)(1) with respect to tt yields the following.

v= v0+at    at= vv0    t= vv0a \begin{align*} && v =&\ v_{ 0 }+at \\ \implies && at =&\ v-v_{ 0 } \\ \implies && t =&\ \frac{ v -v_{0} }{ a } \end{align*}

Substituting it into (2)(2) results in the following.

x= v0vv0a+12a(vv0a)2    x= v0vv0a+12a(vv0)2a2    ax= v0(vv0)+12(vv0)2    2ax= 2v0v2v02+(vv0)2    2ax= 2v0v2v02+v22vv0+v02    2ax= v2v02 \begin{align*} && x =&\ v_{ 0 }\frac { v-v_{0} }{ a }+\frac { 1 }{ 2 }a\left( \frac { v-v_{0} }{ a } \right)^{2} \\ \implies && x =&\ v_{ 0 }\frac { v-v_{0} }{ a }+\frac { 1 }{ 2 }a\frac { \left( v-v_{0} \right)^{2} }{ a^{2} } \\ \implies && ax =&\ v_{ 0 }(v-v_{0})+\frac { 1 }{ 2 }(v-v_{0})^{2} \\ \implies && 2ax =&\ 2v_{0}v-2{v_{0}}^{2}+(v-v_{0})^{2} \\ \implies && 2ax =&\ 2v_{0}v-2{v_{0}}^{2} + v^{2} - 2vv_{0}+v_{0}^{2} \\ \implies && 2ax =&\ v^{2}-v_{0}^{2} \end{align*}

Graphs

The graphs of uniformly accelerated motion are as shown below. A difference from uniform motion graphs is that the graph’s shape changes depending on whether the acceleration is greater or less than 0.

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