Question

An object starts from rest and attains a uniform acceleration of 4m s-1. what will be it's velocity at the end of half meter ?​

Answer 1

v^2 = u^2 +2as

or v^2 = 0 + 2*4*0.5

or v^2 = 4

or v = 2m/s

Answer 2

Answer:

u=0, a=4ms

u=0, a=4ms −2

u=0, a=4ms −2 , s=

u=0, a=4ms −2 , s= 2

u=0, a=4ms −2 , s= 21

u=0, a=4ms −2 , s= 21

u=0, a=4ms −2 , s= 21 m, v=?

u=0, a=4ms −2 , s= 21 m, v=?v

u=0, a=4ms −2 , s= 21 m, v=?v 2

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2as

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4×

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 2

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21 v

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21 v 2

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21 v 2 =4

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21 v 2 =4v=

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21 v 2 =4v= 4

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21 v 2 =4v= 4

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21 v 2 =4v= 4

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21 v 2 =4v= 4 =2ms

u=0, a=4ms −2 , s= 21 m, v=?v 2 =u 2 +2asv 2 =0 2 +2×4× 21 v 2 =4v= 4 =2ms −1