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♠ Derive the newton's laws of motion

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Answer 1

Answer:

the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction

Explanation:

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Answer 2

Question :-

Derive the newton's laws of motion.

Answer :-

Let's learn,

• Firstly we have to know there are three types of Newton's laws of motion.

1. Newton's first law of motion
2. Newton's second law of motion
3. Newton's third law of motion.

• We learn one by one.

Deriving Newton's first law of motion

The first equation of motion can be easily understood by the graph method, for that, let’s look at the graph drawn below:

Let’s suppose that you are walking on the roof of your house at 5 m/s and suddenly the light goes off in your house, you start running out of your house because of fear at a speed of 20 m/s. Now, what is the difference in your velocity? Its is given by:

• v = 20 m/s
• u = 5 m/s,

i.e.,

=> v - u = 20 - 5 = 15 m/s.

So, from the graph: OC - OD = CD = Difference in velocity = 15 m/s.

Now, if you are moving with a uniform acceleration ‘a’, we also know that change in velocity per unit time is acceleration.

From here, a = slope of the graph, which is given as:

• a = $\rm\frac{AB}{AD}$

And, AB = BE - AE

We can see that BE = v and AE = u

Putting the value of AB, we get:

• a = (v - u), and also AD = t

So, we get the formula for acceleration as:

a = change in velocity per unit time = $\rm \frac{v - u}{t}$

Or v - u = at => v = u + at, is the required derivation of the first equation of motion.

If the man was sitting on the roof, which means his u = 0 m/s, then the equation can be re-written as:

• v = at….(1)

Deriving Newton's second law of motion

The second equation of motion describes the Position of an object with respect to time. Let’s derive the second equation of motion with the help of the graph drawn below:

We know that area enclosed by the graph in Fig.a is equal to the distance traveled by the body. So, for finding the distance traveled, we need to find the area of trapezium.

Now, finding the area of trapezium OPQS:

s = Area of Rectangle OPRS + Area of triangle QPR

$\rm{ = (OP \times PR) + \frac{1}{2} \times QR \times PR...(a)}$

Here,

OP = u

PR = t

QR = v - u

On putting these values in equation (a), we get:

$\rm{ = (u \times t ) + \frac{1}{2} \times (v - u) \times t }$

We know that v - u = at

So,

$\rm{s = u \times t + \frac{1}{2} (at) \times t = ut + \frac{1}{2}at {}^{2} }$ , is the required second equation of motion.

Here, s is the distance traveled between two points.

Deriving Newton's third law of motion

The third law of motion describes the velocity-time relationship. Now, let’s derive the third law of motion with the help of Fig.b:

From Fig.b, the area of trapezium can be calculated as:

Area of trapezium OPQS = $\rm{ \frac{1}{2} \times (OP + QS) \times OS = \frac{1}{2} \times (RS + QS) \times OS….(2)}$

Acceleration (a) = slope of velocity-time graph PQ, which is given by:

• a = QR/PR = (QS - RS)/OS

Or

• OS = (QS - RS)/a…..(3)

Now, putting the value of (3) in (2), we get:

= ½ * (RS + QS) * (QS - RS)/a

OS = 1/2a (QS2 - RS2)

Here,

OS = s

RS = u

QS = v

So, we get the third equation of motion as:

s = 1/2a (v² - u²)

Or

2as = (v² - u²)

Or

• v² = u² + 2as

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