Question

Final velocity v according to the third equation of motion ​

Answer 1

Answer:

Step-by-step explanation:

Explanation:

There are 3 equations of motion. They are:

v = u + at

s = ut + 0.5at²

v² = u² + 2as

According to the question, we are required to find velocity in terms of third equation of motion.

Hence considering the third equation we get:

→ v² = u² + 2as

Taking square root on both sides we get:

→ v = √ ( u² +  2as )

Answer 2

Answer: According to third equation of motion : $v^{2} = u^{2} + 2as$

Step-by-step explanation:

Issac Newton gave 3 equations of motion.

He related displacement, velocity, time and acceleration of a body moving with constant acceleration.

The final equation of motion relates final and initial velocity depending on the displacement of particle.

According to third equation of motion :

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

⇒ $v^{2} = u^{2} + 2as$

v ⇒ Final velocity

u ⇒ Initial velocity

a ⇒ acceleration of body

s ⇒ displacement of body

The first two equations of motion are :

$v = u + at$
$s = ut + \frac{1}{2}at^{2}$

Using first two equations of motion, we can derive the third equation of motion

$v = u + at$
⇒ $t = \frac{v-u}{a}$

Substituting the value of t in second equation of motion:

$s = ut + \frac{1}{2}at^{2}$

⇒ $s = u(\frac{v-u}{a}) + \frac{1}{2}a(\frac{v-u}{a} )^{2}$

⇒ $s = \frac{uv-u^{2} }{a} + \frac{v^{2} +u^{2} -2uv}{2a}$

⇒ $s = \frac{2uv-2u^{2} }{2a} + \frac{v^{2} +u^{2} -2uv}{2a}$

⇒  $s = \frac{2uv-2u^{2} +v^{2} +u^{2} -2uv}{2a}$

⇒  $s = \frac{-u^{2} +v^{2} }{2a}$

⇒ $2as = v^{2} -u^{2}$

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