Question

Q3.Draw velocity time graph for a body that has initial velocity ‘u’ and is moving with

uniform acceleration ‘a’. Use it to derive v = u + at; s = ut + ½ at2

, and v

2

= u

2

+ 2a​

Answer 1

~ Firstly according to the graph,

⇢ AO = DC = u (Initial velocity)

⇢ AD = OC = t (Time)

⇢ EO = BC = v (Final velocity)

First equation of motion:

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:

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.

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

Answer 2

Answer:

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