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Question: A particle moving with uniform retardation covers distances 18 m, 14m and 10m in successive seconds....

A particle moving with uniform retardation covers distances 18 m, 14m and 10m in successive seconds. It comes to rest after travelling a further distance of
A. 50m
B. 8 m
C. 12m
D. 42m

Explanation

Solution

The displacement in nth second (Sn)\left( {{S_n}} \right) is given asSn=u+a2(2n−1){S_n} = u + \dfrac{a}{2}\left( {2n - 1} \right). Given that the time is successive let it be x, x+1, x+2 respectively when the displacement is 18m, 14m and 10m respectively. Substituting these values in the equation we can find the retardation (a) and the initial speed (u) .Using the equation of uniform acceleration,v2=u2−2aS{v^2} = {u^2} - 2aS we can find the distance travel (S) before coming to rest.

Complete step by step answer:
Let the initial velocity of the particle be u and its acceleration be aa. We know that the distance travelled in nth second is given as Sn=u+a2(2n−1){S_n} = u + \dfrac{a}{2}\left( {2n - 1} \right).The distance travelled in the 1st second was 18m.
18=u+a218 = u + \dfrac{a}{2} ---------(1)
The distance travelled in 2nd second be 14m.
14=u−a2[2(n+1)−1] ⇒14=u+3a214 = u - \dfrac{a}{2}\left[ {2\left( {n + 1} \right) - 1} \right] \\\ \Rightarrow 14 = u + \dfrac{{3a}}{2} ---------(2)
The distance travelled in the 3rd second was 10m.
10 = u - \dfrac{a}{2}\left[ {2\left( {n + 2} \right) - 1} \right] \\\ \Rightarrow 10 = u - \dfrac{{9a}}{2} \\\ ---------(3)
Subtracting equation 2 from 1, we get
4=a2−3a2=−a ⇒a=−4 4 = \dfrac{a}{2} - \dfrac{{3a}}{2} = - a \\\ \Rightarrow a = - 4 \\\
Substituting the value of a in equation1
18=u−2 ⇒u=20 18 = u - 2 \\\ \Rightarrow u = 20 \\\
Therefore the initial velocity of the particle is 20m/s. The retardation of the particle is 4m/s2m/{s^2}. The equation of uniform acceleration is v2=u2−2aS{v^2} = {u^2} - 2aS.Here the final velocity vv is zero as the body comes to rest.Therefore the distance travelled is given as S=u2−v22aS = \dfrac{{{u^2} - {v^2}}}{{2a}}. Substituting the value of u and a, we get
S=0−2028S = \dfrac{{0 - {{20}^2}}}{8}
S = \dfrac{{400}}{8} \\\ \therefore S= 50\\\
Therefore the distance to be further travelled before the body comes to rest 50−(18+14+10)=8m50-(18+14+10) =8m.

Note: The general formula for a particle moving with uniform acceleration is given as, v2=u2−2aS{v^2} = {u^2} - 2aS where uu is the initial velocity vv is the final velocity, aa is acceleration and SS is the displacement of the particle. Retardation means negative acceleration. Acceleration means by how much amount the velocity of the body increases every second, provided the acceleration is constant. In the same fashion, negative acceleration means by how much the velocity of the body decreases each passing second, provided the acceleration is constant.While solving problems involving equations of motion, we have to keep in mind that vector quantities have directions and so we take the sign accordingly.