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

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)

All the best.
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Sagar Singh
B.Tech, IIT Delhi

```
6 years ago
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)
---------------

```
6 years ago
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