A particle moving with unifrom retardation covers distances 18m, 14m and 10m in successive seconds. It comes to rest after travelling a further distance of

At t = 0, u is the velocity 

Distance travelled in first second = 18m , 

where as , S₁ = U + $\frac{a}{2}(2n - 1)$

                   18 =  u + $\frac{a}{2}$   ----  equation 1.

Distance travelled in second  second = 14m

where as S₂ = U + $\frac{a}{2}(2\times 2-1)$

                  14 = u + $\frac{3a}{2}$  ----  equation 2 

Distance travelled in third second = 10m  

where as S₃ = u - $\frac{a}{2}\left(2(n+2\right)-1)$

                 10 = u - $\frac{9a}{2}$  ----  equation 3

subtracting equation 1 from equation 2 we get, 

        a = -4.

substituting the value of a in equation 1 we get 

    18 = u + $\frac{-4}{2}$

    u = 20 m/sec

Equation for the uniform acceleration is v² = u² - 2as. according to question the body comes to rest , so v = 0.

distance travelled, S = ( u² - v² ) / 2a,   substituting the value of u, a we get 

S = $\frac{0 - {\left(20\right)}^2}{8}$

S = $\frac{400}{8}$

there fore the S = 50 m 

It stops after travelling further distance after the three successive second. 

= 50 - (18+14+10)

= 8 m .

Therefore the final value is 8m.