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bhanuveer danduboyina Grade: 12th Pass
        During an accelerated motion of a particle,what is the relation between the average velocity and the final velocity?

6 years ago

Answers : (3)

SAGAR SINGH - IIT DELHI
879 Points
										

Dear student,




  • Motion with constant acceleration or uniformly accelerated motion is that in which velocity changes at the same rate troughout motion.

  • When the acceleration of the moving object is constant its average acceleration and instantaneous acceleration are equal. Thus we have










    aavg= a = v2-v1
    t2-t1

                                                                                       

  • Let v0 be the velocity at initial time t=0 and v be the velocity of object at some other instant of time say at t2=t then above eq. 7 becomes










    a = v-v0
    t-0

    or , v=v0+at                        

  • Graphically this relation is represented as


  • In the same way average velocity can be written as










    vavg = x-x0
    t-0

    where x0 is the position of object at time t=0 and vavg is the averag velocity between time t=0 to time t.The above equation then gives
    x=x0+vavgt                  
    but for the interval t=0 to t the average velocity is










    vavg = v0+v
    2

                                                             
    Now from eq. 8 we find
    vavg = v0 + ½(at)                  
    putting this in eq. 9 we find
    x = x0 + v0t + ½(at2)
    or,

  • x - x0 = v0t + ½(at2)

























































Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.


All the best.


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6 years ago
vikas askiitian expert
510 Points
										

for constant accleration(a) ....


let initial velocity is u & final velocity  is v then


v2 = u2 + 2aS ...........1                  (S is displacement)


average velocity = total distlacement /total time = S/t       ................2


v = u+at


t = v-u/a      ..............3


Vavg = Sa/(v-u)          ............4


 


from eq 4 & 1 eliminating S


Vavg = v2-u2/2(v-u)


 Vavg = (v+u)/2


 


this is the required relation

6 years ago
Sudheesh Singanamalla
114 Points
										

in a uniformly accelarated motion it is


v = final velocity


a = accelaration


u = initial velocity


t = time taken


s = distance travelled


----------------


so the relation is


v = u + at ------------------------(i)


2*a*s = v2 - u2           ----------------------- (ii)


---------------


 


Please approve

6 years ago
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