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/sAnswer given is 60 m , 100 mMy Answer coming is 60 m , 90 mQ :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?

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