 # During an accelerated motion of a particle,what is the relation between the average velocity and the final velocity? 12 years ago

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)

All the best.

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Sagar Singh

B.Tech, IIT Delhi

12 years ago

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

12 years ago

in a uniformly accelarated motion it is

v = final velocity

a = accelaration

u = initial velocity

t = time taken

s = distance travelled

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so the relation is

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

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

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