# 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..!

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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

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don't know

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what is motion?

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

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

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derive 3rd eq. of motion

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

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HaStS OfFfFfF 2 iT At all THe cOsT !!

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what is aceeleration?it si unit

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derive third equation of motion?prasoon raj

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nice answes

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3 equations of motion

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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..!
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what non sence
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what non sense
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Something is missing here.
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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.

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dtte5rhgyrtyr6tburtutgu
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first on i true

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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

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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!!!

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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..

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v2 - u2 = 2as
s = u+v/2*t
s=u+v/2*v-u/a
s=v2 - u2 / 2a
2as=v2-u2
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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.
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you are land
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We use 3 eq of motion when time is not given .
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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
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Ans 1- (a) seen in this pic

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Ans -1-(b) - seen in this pic

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Ans - 1- (c) - seen in this pic

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It is when it travels in uniform motion or it is at rest.
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Is this right ???

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DIFFERENCE BETWEEN INSTANTANEOUS SPEED AND VELOCITY
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Twinkel ki gand maro
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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..!
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graphical representation

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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
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it is correct but with the graph iam asking friend
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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

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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
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Thanks
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From below picture you can see the best solution for this question.
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It's just a piece of dog poop
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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.
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check page number 107 for the graph and page 108 for the derivation of class 9 science reader
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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 )
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Distance can be equal to u+v/2 or total distance/total time. So, the better way to derive it is this...

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Dear student check this.

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Vkfjckho
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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)
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Hi this. Equation is very easy
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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
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Derive the third equation of motion v2-u2=aals
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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
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Drive second equation of motion
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derive second equation of motion
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0=12+4(m-5) equations how write this question
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See the answer and do it

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Is this you need?

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For thiRD EQUATION OF MOTION

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Cbse fullform name
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V2 _ U2= 2as
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U have to make a velocity time graph then after this calculation will be performed

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May it helps

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It is easy.Go on pg no 108,unit no.8.5.3.You will get the answer.
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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..!
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From first eq. Of motion we know that v=u +at
Now rearrange this as: t=v-u/a. Now from second eq. Of motion we know that s=ut+1/2at^2. Now if u substitute the value of t in above eq. You will get the answer by further simplification
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You can derive by graphical method. shall I derive in graphical method?
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science ka syllabus Hindi
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Kanchana aapko Koi nikal nikal payega
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I hope this is useful for you

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Hope it helps....

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What is the volicity
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The third equation of motion is velocity-displacement equation.

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v^2-u^2=2as
=v^2=u^2+2as
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Derived the third equation

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V2-u2=2as
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From 1st equation of motion,

V = U + AT
Therfore, V - U = AT

We know that

S = (U + V / 2 )×T

∴ S/T = U + V / 2
∴U + V = 2×S/T
∴ U + V = 2S / T
So Substituting eq1 and eq2,

(V - U ) (V + U ) = AT ×2S/T

∴V² - U² = 2AS.

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????
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Hope it is clear to you. Regards
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22222=2as
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Hydrogen
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Distinguish between speed and velocity
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v2=u_4
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Hi
My name is d santu
9th class
K.g.b.school
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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
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v'2 - u'2 = 2as

Final velocity to power 2 minus initial velocity to power 2 is equal to 2 multiplied to acceleration multiplied to distance covered.
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s = speed*time

s = u+ v / 2* t

s = u + v / 2 * v - u / a? { t = v - u / a}
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ijnb
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'''''''''''
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find the distance and displacement about the picture
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I hope you understand ????????

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What
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2as =V2 -u2

We know distance covered= average velocity * time
S= (v+u)/2 *t --- 1
As, t= v-u/a

Substituting 't' in 1 we get-

S= (u+v)/2 * (v-u)/a
S= v^2-u^2/2a
2aS =v^2-u^2

Hence proved..
Hope it helps you
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- 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..!
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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..

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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
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Plz Don't Post Your Photos In This Plz..Plz Don't Post Your Photos In This Plz.
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Graphically
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2as+u2=v2
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it can be derived as :
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 :)
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An object at rest seems to be at motion with respect to another object but for the other object this seems to be at rest so motion and rest are relative terms.
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hope u got the derivation in detail
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Second equation of motion related ....
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• DIFFERENCE BETWEEN INSTANTANEOUS SPEED AND VELOCITY
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distance (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
derivation completed >>>>>>>>>>>>

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a cyclist completes n rounds of a circular track what will be the distance covered and displacement of the cyclist after completing n rounds if D is diameter of the track
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v2- u2as

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  - U= As

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Area of Trapezium =(sum of parallel side ?Height
S=1/2(U+V)?T
S=1/2(U+V)?V-U/a
S=v2-U2
2a
=2as=v2-u2
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Hope this help

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