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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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```										ans1: the ans to displacement is correct.
lets do it for distance. for that you have to find out at which point of time the velocity of particle is changing.
v = u + a*t
0 = 40 +(-10)*t
=> t = 4sec
we will find out the distance travelled by particle in two phases : in first 4 sec and then in last two sec.
in first 4 sec :  s = ut + 0.5a*t^2
= 40 *4 - 0.5*10*4^2
= 80 m
in nxt 2 sec:   now the initial velocity of particle is zero as calculated above, i.e. it is a free fall.
s = ut + 0.5a*t^2
= 0 + 0.5*10*2^2
= 20m
total dis= 80+20 = 100m
ans 2: avg velocity = total displacement / total time
as acc. is contant, a = (v2-v1)/t
s = v1t + 0.5at^2
= v1t + 0.5*(v2-v1)*t^2/t
= v1t + 0.5*(v2-v1)*t
=  0.5(v2+v1)* t
avg vel. = s / t
= 0.5(v2+v1)
ans 3:
s1= u(1)+ 0.5a(1)^2
10 = u + 0.5a      ......(1)
s2= u(2) + 0.5a(2)^2
10 +10= 2u + 2a  ......... (2)   [ the que didn't specify both the distances are in same direction or in opposite direction, i have solved it for same       directions but if the distance is in op directions then substitute 10 - 10 in place of 10+10 n solve, which gives u = 20m/s nd a = -20m/s^2
solving (1) n (2),
we get, u = 10m/s, a =0

```
7 years ago

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