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Vishrant Vasavada Grade: 11
        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.
8 years ago

Answers : (2)

Ramesh V
70 Points
										

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

 

 

8 years ago
AskIITians Expert PRASAD IIT Kharagpur
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.

 


8 years ago
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