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```
A particle starting from rest moves with constant acc. If it takes time t = 5 s to reach the speed 18 km/h, find a) average velocity in this time period and b) distance travelled. (ANSWER GIVEN: a) 2.5 m/s^2 b) 12.5 m) Is it not 1 m/s and 5 m? Justify your ANSWERS.

```
11 years ago

```							consider the equations:
v = u +at
S= ut + at2/2  where u-initial vel
v - final velocity   , a -accln  and S=distance travelled

a) since u=0 , so v=at
given v = 18 km/hr = 5m/sec
so a = v/t = 1 m/sec2
average velocity = ( u + v )/2
so V avg = (0+5) /2 = 2.5               ---------  V avg = 2.5 m/sec
alsoV avg is the area under velocity time curve
b)
a = 1 m/sec2  and t= 5 sec and u=0
S= ut + at2/2
S=1*25/2
S = 12.5 m

```
11 years ago 18 Points
```
initial velocity  u = 0
final velocity v = 18 km/hr = 5 m/s^2.
a = average acceleration.
v = u + a*t      --------- time  t = 5 s
5 = 0 + a*5
a = 1 m/s^2.
Since particle moves in same direction   distance travelled = displacement.
s = u*t + 0.5*a*t^2 = 0 + 0.5*1*25
s = 12.5 m.
average velocity = displacement / time.
Vavg = 12.5 / 5
Vavg = 2.5 m/s.

```
11 years ago
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