Question

derive the equation with the help of the velocity v/s time graph hence obtain v =u + at ​

Answer 1

Explanation:

please mark me as brainlist then you will get the answer .

Answer 2

Answer:

(ㅅꈍ﹃ꈍ)* нєуα мαтє нєяє'ѕ υя αηѕωєя...!!! *(ꈍ﹃ꈍㅅ)♡

Graphical method : Consider an object moving along a straight line with uniform acceleration a let u be the initial velocity of the object at time t=0 and v be the final velocity of the object at time t . let s be the distance travelled by the object in time t.

velocity -time graph of this motion is a straight line PQ as shown in the figure

where OP=u=RS

OW=SQ=v and

OS=PR=t

(I) for the first equation of motion : we know that the slope of velocity -time graph of uniformly accelerated motion represents the acceleration of the object

i.e. acceleration = slope of the velocity -time graph PQ

or a=

PR

QR

​

=

OS

QR

​

a=

OS

SQ−SR

​

=

t

v−u

​

or v−u=at

or v=u+at .....(i)

This is the first equation of uniform acceterated motion.

(ii) Second equation of motion ; We know that the area under the velocity-time graph for a given time interval represents the distance covered by the uniformly accelerated object in that interval of time.

∴ Distance (displacement) travelled by the object in time t is :

S = area of trapezium OSQP

= area of rectangle OSRP + Area of triangle PRQ

or S=OS×OP+

2

1

​

×PR×PQ

(area of rectangle = length × breadth)

( Area of triangle =

2

1

​

× Base × Height )

=t×u+

2

1

​

×t×(v−u)

(from the first equation of motion v−u=at)

=ut+

2

1

​

×t×at

Thus S=ut+

2

1

​

at

2

......(ii)

This is the second equation of uniform accelerated motion.

(iii) Third equation of motion : Distance travelled by the object in time interval t is s = area of trapezium OSQP

=

2

1

​

(OP+SQ)×OS

∵OP=SR

=

2

1

​

(SR+SQ)×OS........(iii)

Acceleration , a = slope of the velocity -time graph PQ

or a=

PR

RQ

​

=

OS

SQ−SR

​

or OS=

a

SQ−SR

​

...........(iv)

Putting this value in equation (iii) we get

s=

2

1

​

(SR+SQ)(

a

SQ−SR

​

)

or s=

2a

1

​

(AQ

2

−SR

2

)

or s=

2a

1

​

(v

2

−u

2

)

or v

2

−u

2

=2as

or v

2

=u

2

+2as.............(v)

This the third equation of uniform accelerated motion.

solution

【ＨＯＰＥ　ＩＴＳ　ＭＡＹ　ＨＥＬＰＦＵＬ　ＴＯ　ＹＯＵ】

【ＭＡＲＫ　ＡＺ　ＢＲＡＩＮＬＩＳＴ　ＡＮＳＷＥＲ．．．　】

【Ｆｏｌｌｏｗ　ｍｅ♦️】

➶➶➶➶➶give thx❣️ ➷➷➷➷➷