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A particle starting from rest moves in a straight line with acceleration proportional to x^2, (where x i displacement). The gain of kinetic energy for any displacement is proportional to : 1) x^3 2)x^1/2 3)x^2/3 4)x^2?
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A particle starting from rest moves in a straight line with accelerati...
**Introduction**

In this problem, we are given that a particle starting from rest moves in a straight line with acceleration proportional to x^2, where x is the displacement. We need to determine the relationship between the gain of kinetic energy and the displacement.

**Derivation**

Let's assume that the acceleration of the particle is given by a = kx^2, where k is a constant of proportionality.

We know that acceleration is the rate of change of velocity with respect to time. Therefore, we can write:

a = dv/dt, where v is the velocity of the particle.

Since the particle starts from rest, we can integrate both sides of the equation to find the velocity as a function of time:

∫a dt = ∫dv

∫kx^2 dt = ∫dv

kt = ∫v dv

kt = (1/2)v^2 + C

where C is the constant of integration.

Now, let's consider the gain of kinetic energy for a displacement x.

We know that kinetic energy is given by KE = (1/2)mv^2, where m is the mass of the particle.

Since the particle starts from rest, we can write the gain of kinetic energy as:

ΔKE = KE_final - KE_initial

ΔKE = (1/2)m(v_final^2 - v_initial^2)

Since the particle starts from rest, the initial velocity is 0. Therefore, the equation simplifies to:

ΔKE = (1/2)mv_final^2

Substituting the expression for velocity obtained earlier, we have:

ΔKE = (1/2)m[(kt - C)^2]

Simplifying further:

ΔKE = (1/2)m(k^2t^2 - 2kCt + C^2)

Now, let's consider the displacement x. We know that velocity is the rate of change of displacement with respect to time. Therefore, we can write:

v = dx/dt

Rearranging the equation, we have:

dt = dx/v

Substituting the expression for velocity obtained earlier, we have:

dt = dx/(kt - C)

Integrating both sides of the equation, we have:

∫dt = ∫dx/(kt - C)

t = (1/k)ln|kt - C| + D

where D is the constant of integration.

Now, let's consider the displacement x for a given time t. We can rearrange the equation obtained earlier to solve for x:

kt - C = e^(ktD)

kt = C + e^(ktD)

x = (1/k)(C + e^(ktD))

where C and D are constants.

**Relationship between Gain of Kinetic Energy and Displacement**

Now, let's substitute the expression for x in terms of t into the equation for ΔKE:

ΔKE = (1/2)m(k^2t^2 - 2kCt + C^2)

ΔKE = (1/2)m(k^2[(1/k)ln|kt - C| + D]^2 - 2kC[(1/k)ln|kt - C| + D] + C^2)

Simplifying further:

ΔKE = (1/2)m(k^2/k^2[ln|kt - C| +
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A particle starting from rest moves in a straight line with acceleration proportional to x^2, (where x i displacement). The gain of kinetic energy for any displacement is proportional to : 1) x^3 2)x^1/2 3)x^2/3 4)x^2?
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A particle starting from rest moves in a straight line with acceleration proportional to x^2, (where x i displacement). The gain of kinetic energy for any displacement is proportional to : 1) x^3 2)x^1/2 3)x^2/3 4)x^2? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A particle starting from rest moves in a straight line with acceleration proportional to x^2, (where x i displacement). The gain of kinetic energy for any displacement is proportional to : 1) x^3 2)x^1/2 3)x^2/3 4)x^2? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle starting from rest moves in a straight line with acceleration proportional to x^2, (where x i displacement). The gain of kinetic energy for any displacement is proportional to : 1) x^3 2)x^1/2 3)x^2/3 4)x^2?.
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