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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 v_{1} and at time t =t is v_{2} . 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?
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 v_{1} and at time t =t is v_{2} . 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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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)
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)
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
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
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....
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....
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