Question

the correctness the equation s+ut+1/2at²​

Answer 1

The correctness of the given equation is mentioned below:

• ${\small{\underline{\boxed{\pmb{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}}$

By using dimensional analysis!

Knowledge required:

${:\implies{\pmb{\sf{\red{Velocity \: = \dfrac{Distance}{Time}}}}}}$

${:\implies{\pmb{\sf{\red{Time \: = \dfrac{Distance}{Velocity}}}}}}$

${:\implies{\pmb{\sf{\red{Distance \: = Velocity \times Time}}}}}$

${:\implies{\pmb{\sf{\red{Acceleration \: = \dfrac{dv}{dt}}}}}}$

Knowledge required too:

→ The dimensional formula of distance or displacement is ${\pmb{\sf{\green{[M^0L^1T^0]}}}}$

→ The dimensional formula of acceleration is ${\pmb{\sf{\green{[M^0L^1T^{-2}]}}}}$

→ The dimensional formula of time is ${\pmb{\sf{\green{[M^0L^0T^1]}}}}$

→ The dimensional formula of velocity is ${\pmb{\sf{\green{[M^0L^1T^{-1}]}}}}$

Now by using dimensional analysis formulas, let's carry on!

Taking left hand side !..

${\sf{:\implies s}}$

${\sf{:\implies [M^0L^1T^0]}}$

Taking right hand side!..

${\sf{:\implies ut \: + \dfrac{1}{2} \: at^2}}$

${\sf{:\implies [M^0L^1T^{-1}] \cdot [M^0L^0T^1] + \dfrac{1}{2} \cdot [M^0L^1T^{-2}] \cdot [M^0L^0T^1] \cdot [M^0L^0T^1]}}$

${\sf{:\implies [M^0L^1T^{-1}] \cdot [M^0L^0T^1] + [M^0L^1T^{-2}] \cdot [M^0L^0T^1] \cdot [M^0L^0T^1]}}$

${\sf{:\implies [M^0L^1T^{-1}] \cdot [M^0L^0T^1] + [M^0L^1T^{-2}] \cdot ([M^0L^0T^1]^{2})}}$

${\sf{:\implies [M^0L^1T^0] + [M^0L^1T^{-2}] \cdot [M^0L^0T^2]}}$

${\sf{:\implies [M^0L^1T^0] + [M^{(0+0)}L^{(0+1)}T^{(-2+2)}]}}$

${\sf{:\implies [M^0L^1T^0] + [M^0L^1T^{0}]}}$

${\sf{:\implies [M^0L^1T^0]}}$

${\pmb{\sf{\orange{[M^0L^1T^0] \: = [M^0L^1T^0]}}}}$

• Henceforth, checked! ✓

____________________________

Additional information:

⋆ There are three equations of motion. The equations of motion are named as

◆ Velocity time relationship

◆ Position time relationship

◆ Position velocity relationship

⋆ Therefore, velocity time relationship, Position time relationship and Position velocity relationship are the three equations of motion respectively.

◆ Velocity time relationship

That's the equation,

• ${\small{\underline{\boxed{\sf{v \: = u \: + at}}}}}$

◆ Position time relationship

That's the equation,

• ${\small{\underline{\boxed{\sf{s \: = ut \: + \dfrac{1}{2} \times at^2}}}}}$

◆ Position velocity relationship

That's the equation,

• ${\small{\underline{\boxed{\sf{2as \: = v^2 - u^2}}}}}$

(Where, v denotes final velocity , u denotes initial velocity , a denotes acceleration , t denotes time , s denotes displacement or distance)

~ Firstly according to the graph,

⇢ AO = DC = u (Initial velocity)

⇢ AD = OC = t (Time)

⇢ EO = BC = v (Final velocity)

Let's derive the 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

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

For better understanding of this answer, you can go to web too! :)

Answer 2

Answer:

LHS:-

S = M⁰L¹T⁰

RHS:-

ut + 1/2at² = [M⁰L¹T-¹][M⁰L⁰T¹] + ½[M⁰L¹T-²][M⁰L⁰T¹]²

(Distance = Speed × Time, Acceleration = Rate of change of velocity.)

= M⁰L¹T⁰ + M⁰L⁰T⁰

= M⁰L¹T⁰ = RHS.

Therefore, verified.

More: Checking the correctness of an equation through dimension analysis is not enough.