# derive the third equation of motion : v2 - u2 =2as

v2- u2 = 2as

since, S (Distance) = Average speed x Time

S = U+V / 2 * T

S = U+V / 2 * V - U / A   {since T = V -U / A}

S = V2 - U2 / 2A

2AS = V2 - U2

OR  V2  - U2 = As

Hence,Derived..!

• 56

v2-u2=2as

s = speed*time

s = u+ v / 2* t

s = u + v / 2 * v - u / a  { t = v - u / a}

s = v 2- u 2 /2 a

2as= v2 - u2

hence drived

• 28

don't know

• -5

what is motion?

• 0

Motion is the change of postion of an obejct with respect to any stationary object (object at rest).

• 2 Click thsi for solution...
The first one is also right

• -1

derive 3rd eq. of motion

• 1

derive the third equation of motion : v2 - u2 =2as by graphical method

• -4

I LOveD It It waS AwSzmM Nd hELpEd mE OuT SoLviNg thE SUm Of pHySiCs
HaStS OfFfFfF 2 iT At all THe cOsT !!

• -4

what is aceeleration?it si unit

• 2

derive third equation of motion?prasoon raj

• -1

nice answes

• -1
3 equations of motion

• 1
v2- u2 = 2as since, S (Distance) = Average speed x Time S = U+V / 2 * T S = U+V / 2 * V - U / A {since T = V -U / A} S = V2 - U2 / 2A 2AS = V2 - U2 OR V2 - U2 = As Hence,Derived..!
• 1
what non sence
• -7
what non sense
• -11
Something is missing here.
• 4
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.

• 18
dtte5rhgyrtyr6tburtutgu
• -19
first on i true

• -5
2as=v2-u2
Did u guys thought of doing it in the opposite way
2as=v2-u2
as=v2-u2/2
s=(v+u)(v-u)/2a
and so on

• 4
v2=u2+2as
v=u+at
v2=(u+at)2             (squaring both sides)
v2=u2+2uat+at2
v2=u2+2a​ (ut+1/2 at2)
v2=u2+2as              (s= ut+1/2 at2)
Hence, derived.
Hope this helps you!!!

• -1
For this equation we need 2 equations as base which are : s = v+u/2 * time and t = v-u/a
s(average velocity)= v + u/2 * time

In the equation we don't want  time, so we will use the equation t= v-u/a derived from 1st equation of motion
=> s = v+u/2*v-u/a
=> s = (v+u)(v-u)/ 2*a
=> s = v^2 - u^2 / 2a
​=> 2as = v^2 - u^2
=> 2as + u^2 = v^2
Hence proved..

• 1  • 0  • 3  • 0
v2 - u2 = 2as
s = u+v/2*t
s=u+v/2*v-u/a
s=v2 - u2 / 2a
2as=v2-u2
• 4
v2- u2 = 2as

since, S (Distance) = Average speed x Time

? S = U+V / 2 * T

? S = U+V / 2 * V - U / A ? {since T = V -U / A}

? S = V2 - U2 / 2A

? 2AS = V2 - U2

? OR? V2? - U2 = As

Hence,Derived..! I hope it was useful for all of you.
• 0
you are land
• -4
We use 3 eq of motion when time is not given .
• 1 • -3
Ans 1- Physical quantity whose SI unit is m/sec is speed (or velocity). Ans 2 - Physical qunatity whose SI unit is m/sec square is Acceleration
• -3 • -2
Ans 1- (a) seen in this pic • 2
Ans -1-(b) - seen in this pic • -3
Ans - 1- (c) - seen in this pic • -2
It is when it travels in uniform motion or it is at rest.
• 0
Is this right ??? • -1
DIFFERENCE BETWEEN INSTANTANEOUS SPEED AND VELOCITY
• -6
Twinkel ki gand maro
• -9
v2- u2?=?2as

since, S (Distance) = Average speed x Time

? S = U+V / 2 * T

? S = U+V / 2 * V - U / A ? {since T = V -U / A}

? S = V2?- U2?/ 2A

? 2AS = V2?- U2

? OR? V2? -?U2?= As

Hence,Derived..!
• 8
graphical representation • 3
Uniform Motions:- 1.The hour hand of a clock - It moves with uniform speed, completing movement of a specific distance in an hour 2.A car going along a straight level road at steady speed 3.An aircraft cruising at a level height and a steady speed 4.A ship steaming on a straight course at steady speed 5.A train going along the tracks at steady speed 6.A cooling fan running at a fixed speed 7.Earth moving round the sun is an uniform motion 8. Movement of fan 9. A pendulum having equal amplitudes on both sides 10. A vibrating spring in a sewing machine 11. Rain drops fall at uniform speed as buoyant forces balances 'g' Non uniform Motions:- 1. A horse running in a race 2. A bus on its way through the market 3. A bouncing ball 4. Movement of an asteroid 5. Aircraft moving through the clouds and then landing 6. Dragging a box from a path 7. A man running a 100 m race 8. A car coming to a halt 9. A train coming to its terminating sop 10. A car colliding with another car In the above motions, the speed of varying
• -2
it is correct but with the graph iam asking friend
• -6
v2-u2=2as

s = speed*time

s = u+ v / 2* t

s = u + v / 2 * v - u / a? { t = v - u / a}

s = v 2- u 2 /2 a

2as= v2 - u2

hence drived

• -2
first you must find the area of trapezium USING THE FORMULA
so,
AREA OF TRAPEZIUM=1/2 X t X (v+u)

=1/2(v - u/2)(v+u)

S  = {(v - u)(v+u)}/

2 a

2       2
2 A S =V  -  U
• 6
Thanks
• -2
From below picture you can see the best solution for this question.
• -2
It's just a piece of dog poop
• -1 • -3
When the speed of an obj is constantly changing,the instantaneous speed is the speed of an obj at a particular moment in time. For ex- A cheetah who is runningwith the speed of 80 mins per hour then it is shown as in per hour speed. The velocity of an obj in motion at a specific point in time. If an obj has a standard velocity over a period of time its average and instantaneous velocity.
• -1
check page number 107 for the graph and page 108 for the derivation of class 9 science reader
• 0
Since, the third equation of motion is given as hereunder:-
V?-U?=2as where
V=final velocity
U=initial velocity
a=Acceleration
s= Displacement
So the derivation of equation is given hereunder:-
Since, Displacement= average velocity ?Time
So, S= v+u?2 ? v-u?a(acceleration)
S=V?-U??2a
So , V?-U?=2as ( by transvering )
• -1
Distance can be equal to u+v/2 or total distance/total time. So, the better way to derive it is this... • 4 • 0 • 0 • 0
Dear student check this. • -1 • 2
Vkfjckho
• 0 • 0 • 0
S=avg.v×t S=t(u+v)÷2 Using T=(v-u)÷a S=(v-u)÷a×(v+a)÷2 S=(v^2-u^2)÷2a 2as=(v^2-u^2)
• 0
Hi this. Equation is very easy
• 0
Physicalquantitieswhich need only magnitude to define it completelyare known as scalar quantities ad physical quantities which need both magnitude and direction to define it completely are known as vector quantities
• 0
Derive the third equation of motion v2-u2=aals
• 0
From the velocity - time graph the distance 's' covered by an object in time(t) moving with uniform accelerated motion is the area enclosed by the trapezium OABC under the graph.

Distance(s) = Area of trapezium OABC
s =1/2 (OA +BC)OC
s =1/2 (u+v)t
s = t = v - u / a
s = 1/2 (v+u) (v-u/a)
2as =(v+u) (v-u)
2 2
2as = v - u
• 0 • 0 • 0 • 0 • 2
Drive second equation of motion
• 0
derive second equation of motion
• 0 • 0
0=12+4(m-5) equations how write this question
• 0 • 0 • 0 • 0 • 0 • 0 • 0 • 0 • 0
See the answer and do it • 0 • 0 • 0
Is this you need? • 0 • 0 • 0 • 0
For thiRD EQUATION OF MOTION • 0
Cbse fullform name
• 0 • 0 • 0 • 0
V2 _ U2= 2as
• 0
U have to make a velocity time graph then after this calculation will be performed • 0 • 0 • 0
May it helps • 0
It is easy.Go on pg no 108,unit no.8.5.3.You will get the answer.
• 0
v2- u2?=?2as

since, S (Distance) = Average speed x Time

? S = U+V / 2 * T

? S = U+V / 2 * V - U / A ? {since T = V -U / A}

? S = V2?- U2?/ 2A

? 2AS = V2?- U2

? OR? V2? -?U2?= As

Hence,Derived..!
• 0
What are you looking for?