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A particle starting from rest and moving under constant force displaces by S1 in time t. It shall displace further by distance S2 in time from t to 2t, then what is true,
1. S2= 2s1
2. S2 = 3s1
3. S2 = 4s1
4. \(s_2 = \frac{s_1}{2}\)

1 Answer

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Best answer
Correct Answer - Option 2 : S2 = 3s1

CONCEPT:

  • Equation of Kinematics: These are the various relations between u, v, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
  • Equations of motion can be written as

V = U + at

\(s =ut+\frac{1}{2}{at^{2}}\)

V2 =U2+ 2as

where, U = Initial velocity, V = Final velocity, g = Acceleration due to gravity, t = time, and h = height/Distance covered

CALCULATION:

Given Initial velocity u = 0 m/s

  • We know that when the force is constant, the acceleration is also constant.

The second equation of motion is given as,

\(⇒ S=ut+\frac{1}{2}at^{2}\)   

For time t,

⇒ displacement S = S1,

\(⇒ S_{1}=(0\times t)+\frac{1}{2}at^{2}\)

\(⇒ S_{1}=\frac{1}{2}at^{2}\)   ------- (1)

For total time 2t,

⇒ displacement S = S1+S2,

\(⇒ S_{1}+S_{2}=(0\times 2t)+\frac{1}{2}a(2t)^{2}\)

\(⇒ S_{1}+S_{2}=4\frac{1}{2}at^{2}\)  ------- (2)

By equating equation 1 and equation 2, we get

\(⇒ S_{1}+S_{2}=4S_{1}\)

\(⇒S_{2}=3S_{1}\)

  • Hence, option 2 is correct.

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