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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.
7 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


 


 

7 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.


 




7 years ago
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