Question

drive the equation bof motion x = ut + ½at² vel​

Answer 1

Correct Question :-

• Drive the equation of motion s = ut + ½ at²

To prove :-

• s = ut + ½ at²

Proof :-

As we know that

• Average velocity = u + v/2

Where " u " is initial velocity and " v " is final velocity.

Distance travelled = Average velocity × time

→ s = (u + v)/2 × t

According to first equation of motion

→ v = u + at

So, put the value of " v "

→ s = (u + u + at)/2 × t

→ s = (2u + at)/2 × t

→ s = 2ut/2 + at²/2

→ s = ut + ½ at²

• Hence proved

Answer 2

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: Correct Question :- \: \star}}}}}$

Derive the second equation of motion :-

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

$\red{\bigstar}$ Second equation of motion

$\huge\leadsto{\sf\purple{Derive : }}$

Let us consider, an object is moving along a straight line with a uniform acceleration (a) at t = 0 , initial velocity (u) . Let s be the distance travelled by the object in time t , from A to b.

↠ s = area enclosed by velocity time graph and time axis.

↠ s = Area of OABC

↠ s = Area of OACD + Area of ∆ ABD.

↠ s = OA × AD + 1/2 × BD × AD

↠ s = OA × OC + 1/2 (BC - CD) × OC. ...Eq(1)

OA = u

OC = t

BD = BC - CD = v - u.

↠ s = u × t + 1/2 (v - u) t

From first equation of motion we have , v - u = at.

↠ s = u × t × 1/2 × at × t

$\therefore\underline{\boxed{\textsf{s = {\textbf{ut + 1/2at²}}}}} \qquad\qquad \bigg\lgroup\bold{2nd \ equation \ of \ motion} \bigg\rgroup$

Hence ,

Derived !

Required information :-

You can derive first equation also in the same way .

✮ First Equation of Motion

Let us consider an object is moving with uniform acceleration along a straight line . Let (u) be the initial velocity at t = 0 . After interval of time (t) , its velocity becomes u.

↠ Let OA = u [ t = 0 ]

Here in the figure we draw AD ⟂ BC , and BE ⟂ OY.

↠ ∠BAD = ϴ. [ say ]

Now,

↠ Acceleration = slope of Line AB

↠ a = tan ϴ

↠ a = BD/AD. [ tan ϴ = P/B ]

↠ a = (v - u)/t. [ AD = OC = t ]

$\therefore\underline{\boxed{\textsf{v = {\textbf{u + at}}}}} \qquad\qquad \bigg\lgroup\bold{1st \ equation \ of \ motion} \bigg\rgroup$

Or

$\underline{\boxed{\textsf{at = {\textbf{v - u}}}}} \qquad\qquad \bigg\lgroup\bold{1st \ equation \ of \ motion} \bigg\rgroup$

More related information :-

✮ First equation of motion

❥ v = u + at.

v = final velocity.

u = initial velocity.

a = acceleration.

t = time taken.

Note :

first equation of motion also called velocity time equation.

✮ Second equation of motion

↠ s = ut + ½ at².

u = initial velocity.

a = acceleration.

t = time taken.

s = distance covered.

Note :-

Second equation of motion also called positive time equation.

✮ Third equation of motion

❥ v² - u² = 2as

v = final velocity.

u = initial velocity.

a = acceleration.

Note :-

Third equation of motion also called velocity positive equation.

❥ First law of motion

↠ first law of motion states that an object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force.

❥ Second law of motion

↠ second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

❥ Third law of motion

↠ third law of motion states that for every action there is an equal and opposite reaction and they act on two different bodies.