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NEET 2009
Exam
NEET
Year
2009
Subject
Physics
Chapter
Motion in One Dimension

NEET 2009 · Physics · Motion in One Dimension

A particle starts its motion from rest under the action of a...

A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 s10 \mathrm{~s} is s1\mathrm{s}_1 and that covered in the first 20 s20 \mathrm{~s} is s2\mathrm{s}_2, then

Options

  1. As2=2 s1\mathrm{s}_2=2 \mathrm{~s}_1
  2. Bs2=3 s1\mathrm{s}_2=3 \mathrm{~s}_1
  3. Cs2=4s1s_2=4 s_1
  4. Ds2=s1s_2=s_1

Correct answer

C. s2=4s1s_2=4 s_1

Step-by-step solution

Key Idea If the particle is moving in a straight line under the action of a constant force then distance covered s=ut+12\mathrm{s}=\mathrm{ut}+\frac{1}{2} at 2^2. Since the body start from rest u=0∴s=12at2\mathrm{u}=0\therefore \quad \mathrm{s}=\frac{1}{2} \mathrm{at}^2 Now, s1=12a(10)2\quad \mathrm{s}_1=\frac{1}{2} \mathrm{a}(10)^2 and s2=12a(20)2DividingEq.(i)andEq.(ii),weget\mathrm{s}_{2}=\frac{1}{2} \mathrm{a}(20)^2 Dividing Eq. (i) and Eq. (ii), we gets1 s2=(10)2(20)2⇒s2=4 s1\begin{aligned} \frac{\mathrm{s}_1}{\mathrm{~s}_2}& =\frac{(10)^2}{(20)^2}\\ \Rightarrow \quad s_2 & =4 \mathrm{~s}_{1} \end{aligned}$
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