Question

derivation of the equation of the motion in class 9th physics​

Answer 1

Answer:

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Explanation:

Ans.

(1) First equation of Motion:

V = u + at

soln.

Consider a body of mass “m” having initial velocity “u”.Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”.

Now we know that:

Acceleration = change in velocity/Time taken

=> Acceleration = Final velocity-Initial velocity / time taken

=> a = v-u /t

=>at = v-u

or v = u + at

This is the first equation of motion.

—————————————-

(2) Second equation of motion:

s = ut + 1/2 at^2

sol.

Let the distance travelled by the body be “s”.

We know that

Distance = Average velocity X Time

Also, Average velocity = (u+v)/2

.: Distance (t) = (u+v)/2 X t …….eq.(1)

Again we know that:

v = u + at

substituting this value of “v” in eq.(2), we get

s = (u+u+at)/2 x t

=>s = (2u+at)/2 X t

=>s = (2ut+at^2)/2

=>s = 2ut/2 + at^2/2

or s = ut +1/2 at^2

This is the 2nd equation of motion.

……………………………………………………………

(3) Third equation of Motion

v^2 = u^2 +2as

sol.

We know that

V = u + at

=> v-u = at

or t = (v-u)/a ………..eq.(3)

Also we know that

Distance = average velocity X Time

.: s = [(v+u)/2] X [(v-u)/a]

=> s = (v^2 – u^2)/2a

=>2as = v^2 – u^2

or v^2 = u^2 + 2as

This is the third equation of motion.

Answer 2

Answer:

Answer:

$\sf Hlo \: mate \: here's \: your \: answer \: Hope \: it \: helps!$

$\huge \sf Equations \: of \: Motion$

$\sf Consider \: a \: body \: moving \: with \: initial \: velocity \: 'u' \: changes \: it's \: velocity \: to$

$\sf 'v' \: after \: 't' \: seconds. \: Let \: 's' \: is \: the \: displacement \: and \: 'a' \: is \: the$

$\sf acceleration \: of \: the \: body.$

From the defenition of acceleration

a = Change in velocity / time

= (v- u )/ t

by cross multiplication

. at = v - u

$\sf \implies v = u + at$

2) displacement (s) = average velocity × time

S = (u + v /2) × t

= [u + (u + at )] / 2 ×t

=( 2u + at ) / 2. × t

= 2ut/2 + 1/2 × ${at}^{2}$

$\sf \implies S \: = \: ut \: + \: \frac{1} {2} {at}^{2}$

3). v = u + at

by squaring on both sides

$\sf {v}^{2} \: = \: ({u \: + \: at})^{2}$

= $\sf {u}^{2} \: + \: 2uat \: + \: {a}^{2} {t}^{2}$

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

Hope it helps you.....