• # question_answer A particle starts its motion from rest under the   action of a constant force. If the distance covered in first 10 s is ${{s}_{1}}$and that covered in the first 20 s is ${{s}_{2}},$then A) ${{s}_{2}}=2{{s}_{1}}$B) ${{s}_{2}}=3{{s}_{1}}$C) ${{s}_{2}}=4{{s}_{1}}$D) ${{s}_{2}}={{s}_{1}}$

If the particle is moving in a straight line under the action of a constant force or under constant acceleration [a] Using, $s=ut+\frac{1}{2}a{{t}^{2}}.$ Since the body starts from the rest$u=0$ $\therefore$ $s=\frac{1}{2}a{{t}^{2}}$ Now, ${{s}_{1}}=\frac{1}{2}a{{(10)}^{2}}$ ?(i) and ${{s}_{2}}=\frac{1}{2}a{{(20)}^{2}}$ ?(ii) Dividing Eq. (i) and Eq. (ii), we get $\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{{{(10)}^{2}}}{{{(20)}^{2}}}\Rightarrow {{s}_{2}}=4{{s}_{1}}$