Question

. Derive the equations of motion using graph.

V = U + a t

S = U t + ½ at2

V

2

- U

2

= 2aS​

Answer 1

Solution:

According to the graph,

⇢ AO = DC = u (Initial velocity)

⇢ AD = OC = t (Time)

⇢ EO = BC = v (Final velocity)

Velocity-time relation:

Firstly we can write BC as BD + DC. Now as BD have the velocity position and DC have the time position. Henceforth, we already know that

$\begin{gathered}\tt \Rightarrow Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ \tt \Rightarrow a \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ \tt \Rightarrow a \: = \dfrac{v-u}{t} \\ \\ \tt \Rightarrow a \: = \dfrac{BD}{t} \\ \\ \tt \Rightarrow at \: = BD\end{gathered}$

Position-time relation:

Firstly let, the object is travelling a distance in time under uniform acceleration. Now according to the graph we are able to see that inside the graph the obtained area enclose within OABC(trapezium) under velocity time graph AB. Therefore, distance travelled by an object can be given by

$\tt \Rightarrow Distance \: = Area \: enclosed \\ \\ \tt \Rightarrow s \: = Area \: enclosed \\ \\ \tt \Rightarrow s \: = OABC \\ \\ \tt \Rightarrow s \: = Area \: of \: rectangle \: + Area \: of \: \triangle \\ \\ \tt \Rightarrow s \: = Length \times Breadth + \dfrac{1}{2} \times Base \times Height \\ \\ \tt \Rightarrow s \: = AO \times AD + \dfrac{1}{2} \times AD \times BD \\ \\ \tt \Rightarrow s \: = u \times t + \dfrac{1}{2} \times t \times BD \\ \\ \tt \Rightarrow s \: = u \times t + \dfrac{1}{2} \times t \times at \\ \\ \tt \Rightarrow s \: = ut + \dfrac{1}{2} \times at^2 \\ \\ {\pmb{\sf{Henceforth, \: derived!}}}$

Third equation of motion:

⇢ From the velocity-time graph shown in the attachment the distance s travelled by the object in time t moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph.

$\begin{gathered}:\implies \tt Distance \: = OABC \\ \\ :\implies \tt Distance \: = OABC \: trapeizum \: area \\ \\ :\implies \tt s \: = OABC \: trapeizum \: area \\ \\ :\implies \tt Distance \: = \dfrac{1}{2} \times (Sum \: of \: parallel \: sides) \times Height \\ \\ :\implies \tt s \: = \dfrac{1}{2} \times (a+b) \times h \\ \\ :\implies \tt s \: = \dfrac{1}{2} \times (AO+BC) \times OC \\ \\ :\implies \tt s \: = \dfrac{1}{2} \times (u+v) \times t \\ \\ :\implies \tt s \: = \dfrac{1}{2} \times (u+vt) \\ \\ :\implies \tt s \: = \dfrac{u+vt}{2} \quad {\pmb{\sf{Eq_n \: 1^{st}}}} \\ \\ \sf From \: v-t \: relationship \: we \: get \\ \\ :\implies \tt t \: = \dfrac{v-u}{a} \quad {\pmb{\sf{Eq_n \: 2^{nd}}}} \\ \\ \sf From \: 1st \: \& \: 2nd \: equation \\ \\ :\implies \tt s \: = \dfrac{(v+u) \times (v-u)}{2a} \\ \\ :\implies \tt s \: = \dfrac{v^2 - u^2}{2a} \\ \\ :\implies \tt 2as \: = v^2 - u^2 \\ \\ {\pmb{\sf{Henceforth, \: derived!!!}}}\end{gathered}$