Question

DERIVE THE THREE EQUATIONS OF MOTION GRAPHICALLY




please do it in graph and send
grade 9​

Answer 1

According to the graph,

⇢ AO = DC = u (Initial velocity)

⇢ AD = OC = t (Time)

⇢ EO = BC = v (Final velocity)

First equation of motion:

Explanation:

Considering an object that moves under uniform acceleration where u is not equal to zero from the graph we are able to see that the initial velocity of the object that is at point A and then it increased to final values that is at point B in the time, the velocity change at the uniform rate acceleration. In the graph the perpendicular line BC and BE are drawn from point B on the time and validity axis respectively.

Now let us derive velocity-time relation. Let's do it.

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

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

As we write BD at the place of v-u henceforth,

$\tt \Rightarrow v - u \: = at \\ \\ \tt \Rightarrow v = \: u \: + at \\ \\ {\pmb{\sf{Henceforth, \: derived!}}}$

Second equation of motion:

Now let us derive 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-tme 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.

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