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Q ; A particle is projected vertically upward with an initial velocity 40 m/s. Find the displacement and distance covered by particle in 6 seconds . Take g = 10 m/s


Answer given is 60 m  , 100 m


My Answer coming is 60 m , 90 m


 


Q :A particle moves in straight line with uniform acc. its vel. at time t=0 is v1 and at time t =t is v2 . the avg vel. of particle in this time interval is ?


 


Q:A particle moves in a straight line with const. acc..If it covers 10 m in first second and extra 10 m in next second.Find its initial velocity?


7 years ago

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Answers : (3)

										

Dear Shobhit,


This is a very straight forward question


at t = 0, u=40


since gravity is in the opposite direction as of velocity, it will retard the motion and a time will come when its upward velocity will become 0.


to calculate that time, use 1st equation of motion


v = u + a*t,    a = -10 ,


0 = 40 - 10 * t


=> t = 4 s


at t = 4 s particle will start coming down


Distance covered till 4 s ,


v^2 = u^2 + 2* a * s


s= 40*40/2*10


s= 80 m


from t=4 to t=6 , it will come down


s1=0.5*a* t^2


s1=0.5*10*2*2


s1=20m


Displacement = 80-20 = 60m


Distance = 80 + 20 = 100m(as it covers 20m coming down)


 


 

7 years ago
										

For Q2


avg. vel.= (v1+v2)/2


since acc is uniform its avg. vel can be calculated very easily


 


for Q3


Let its initial velocity = u


and acc.= a


velocity after 1 sec = u + a


now


10= u + 1/2*a*1^2(after 1 s)


for next sec,


20 = u+1/2*a*2^2 


Taking difference of these equations, we get


a = 20/3


and putting this value of a in 1st eqn , we get u=20/3 m/s


 

7 years ago
										

for q1. jst cheqout aftr hw mch tym dus d vel bcum 0:


given u=40


v=0; g=10


using v=u-gt


u get t=4 sec.......i.e.aftr 4 sec d vel bcums zero n d tym givn in ques is 6 sec'wich means dat d particl went up and aftr 4 sec it cam to rest den again gaind acc n came down...........so displacement can b calc. using s=ut-0.5g(t*t)


d dist covrd will be (dist cov b4 cumin 2 rest ) + (dist cov in 6-4=2 sec)........


n u'll gt d rite ans...


 


q2.    Avg vel=net displacement divided by total time taken


acc=(v2-v1)/t


u=v1


v=v2


calc 's'


total tym is t.....


hence calc avg vel.


 


Q3.u cn solve d ques using 2 eq.


for 1st eq.


s=10; t=1; u and a unknown;


s=ut+0.5a(t*t)


for  2nd eq.


s=10; t=1; a remains same and u=vel attaind aftr 1 sec i.e.(u+at) and t=1


 


solve d 2 eq. simultaneously n u'll gt u and v both....


 


 

7 years ago

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