During an accelerated motion of a particle,what is the relation between the average velocity and the final velocity?

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 v_avg is the averag velocity between time t=0 to time t.The above equation then gives
  x=x₀+v_avgt                  
  but for the interval t=0 to t the average velocity is

  vavg =	v0+v
  	2

                                                           
  Now from eq. 8 we find
  v_avg = 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.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi         

Approved

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

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

from eq 4 & 1 eliminating S

V_avg = v²-u²/2(v-u)

 V_avg = (v+u)/2

this is the required relation

Approved

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

Approved