Question

Starting from rest, a body travels $36m$ in the first $2s$ of its journey. The distance it can travel in the ${11^{th}}$ second is
(A) $72m$
(B) $108m$
(C) $144m$
(D) $189m$

Answer 1

Hint: In order to solve this question, we will first calculate the acceleration of the body using given conditions and then we will subtract the distance covered by the body between $11$ seconds and $10$ seconds which will be the actual distance covered by the body in ${11^{th}}$ second.

Formula used: Newton’s equation of motion is given as $S = ut + \dfrac{1}{2}a{t^2}$ where,
S is the distance covered by the body with initial velocity u, acceleration a and in time t.

Complete answer:
We have given that, initial velocity of the body is $u = 0$ distance covered in first $t = 2s$ is $S = 36m$ and let acceleration of the body is a then using formula $S = ut + \dfrac{1}{2}a{t^2}$ and putting the values we get,
 $
  36 = 0(2) + \dfrac{1}{2}a{(2)^2} \\
  36 = 2a \\
  a = 18m{s^{ - 2}} \to (i) \\
 $
Now, the distance during ${11^{th}}$ second will be difference between distance covered by the body between $11$ seconds and $10$ seconds which can be written as
${S_{t = 11}} - {S_{t = 10}}$ so, using acceleration value from equation (i) and $t = 11$ we get
$
  {S_{t = 11}} = 0 + \dfrac{1}{2}(18){(11)^2} \\
  {S_{t = 11}} = \dfrac{1}{2}(18){(11)^2} \to (ii) \\
 $
Similarly for $t = 10$ we get,
$
  {S_{t = 10}} = 0 + \dfrac{1}{2}(18){(10)^2} \\
  {S_{t = 10}} = \dfrac{1}{2}(18){(10)^2} \to (iii) \\
 $
Now, subtracting equation (iii) from equation (ii) we get,
$
  {S_{t = 11}} - {S_{t = 10}} = \dfrac{1}{2}18{(11)^2} - \dfrac{1}{2}18{(10)^2} \\
  {S_{t = 11}} - {S_{t = 10}} = 9(121 - 100) \\
  {S_{t = 11}} - {S_{t = 10}} = 189m \\
 $
So, the distance covered at ${11^{th}}$ second is $189m$

Hence, the correct answer is option (D) $189m$

Note: It should be remembered that, when a body starts from rest its initial velocity is always taken as zero and the other two most important newton’s equations of motion that are used to solve such problems are $v = u + at$ and ${v^2} - {u^2} = 2aS$ where v is the final velocity of the body at time t, These kinds of all kinematics problem can be solved by using these three newton’s equation of motion.