Question

prove
${v}^{2} = {4}^{2} + 2as$
​

Answer 1

Question : -

Prove that v² = u² + 2as ?

ANSWER

Required to prove : -

The 3rd Equation of motion !

• v² = u² + 2as

Proof : -

There are 3 equations of motion;

• v = u+at
• s = ut+½at²
• v² = u²+2as

Here,

v = final velocity

u = initial velocity

s = displacement

t = time taken

a = acceleration

Each and every equation of motion will have some or the other relation !

So,

Let's explore about what relations will they have in between them .

1st equation gives us the relation between the final velocity (v) and time taken (t).

2nd equation gives us the relation between the displacement (s) and time taken (t).

3rd equation gives us the relation between the final velocity (v) and displacement (s).

Now,

Consider the 2nd Equation of motion !

»s = ut+½at² .....(1)

In the 3rd equation of motion , the variable 't' will not be present. So, let's manipulate the variable 't' in the 2nd equation.

From the 1st equation

v = u+at

v-u = at

t = (v-u)/(a) ......(2)

Substituting the value of 't' in equation 1

s = u[(v-u)/(a)]+½[a][{(v-u)/(a)}²]

s = (uv-u²)/(a)+(a)/(2)[(v²+u²-2uv)/a²]

s = (uv-u²)/(a)+(a[v²+u²-2uv])/(2a²)

s = (2a[uv-u²]+a[v²+u²-2uv])/(2a²)

s = (2auv-2au²+av²+au²-2auv)/(2a²)

s = (av²-au²)/(2a²)

s = (a[v²-u²])/(2a[a])

Cancelling 'a' in both numerator and denominator on the RHS part

s = (v²-u²)/(2a)

By cross multiplication

2as = v²-u²

u²+2as = v²

• ⟩⟩⟩ v² = u²+2as ★

Hence Proved !

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

• Prove v² = u² + 2as

★═════════════════★

♣ ᴛᴏ ᴘʀᴏᴠᴇ :

v² = u² + 2as

★═════════════════★

♣ ᴘʀᴏᴏꜰ :

We know :

First Equation of Motion : v = u + at

Second Equation of Motion : S = ut + ½ at²

Third Equation of Motion : v² = u² + 2as

Make "t (time taken)" as subject :

$\sf{t=\dfrac{v-u}{a}}$

Substitute value of "t" in S = ut + ½ at² (Second Equation of Motion) :

$\sf{\mathrm{S}=\mathrm{u}\left(\dfrac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}\right)+\dfrac{1}{2} \mathrm{a}\left(\dfrac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}\right)^{2}}$

Multiplying Both sides by 2a :

$\sf{\mathrm{2aS}=2\left(\mathrm{u}\left(\dfrac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}\right)+\dfrac{1}{2} \mathrm{a}\left(\dfrac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}\right)^{2}\right)}$

$\displaystyle2aS=2a\left(u\frac{v-u}{a}+\frac{1}{2}a\left(\frac{v-u}{a}\right)^2\right)$

$\displaystyle2aS=2a\left(\frac{u\left(v-u\right)}{a}+\frac{\left(v-u\right)^2}{2a}\right)$

$2aS=2a\dfrac{v^2-u^2}{2a}$

$2aS=\dfrac{\left(-u^2+v^2\right)\cdot \:2a}{2a}$

$2aS=-\dfrac{\left(-u^2+v^2\right)a}{a}$

$\sf{2aS=-u^2+v^2}$

$\boxed{\sf{2aS=v^2-u^2}}$

But, we need to prove v² = u² + 2as :

Let's solve for v² in 2aS = v² - u²

$\sf{2as=v^2-u^2}$

$\implies\sf{v^2-u^2=2as}$

$\implies\sf{v^2-u^2+u^2=2as+u^2}$

$\huge\boxed{\sf{v^2=2as+u^2}}$

Hence Proved !!!

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