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Motion in A Straight Line question

2010 · Q165
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Motion in A Straight Line question

2010 · Q165

NEETPhysicsMotion in A Straight LineMCQ+4 / −1
A particle moves a distance x in time t according to equation x = (t + 5)−-−1. The acceleration of particle is proportional to
  1. A
    (velocity)3/2
  2. B
    (distance)2
  3. C
    (distance)−-−2
  4. D
    (velocity)2/3
View written solutionFree

Correct answer: A

x = 1t+5{1 \over {t + 5}}t+51​

∴\therefore∴ v = dxdt{{dx} \over {dt}}dtdx​ = −1(t+5)2{{ - 1} \over {{{\left( {t + 5} \right)}^2}}}(t+5)2−1​ .......(i)

∴\therefore∴ aaa = dvdt{{dv} \over {dt}}dtdv​ = 2(t+5)3{2 \over {{{\left( {t + 5} \right)}^3}}}(t+5)32​ = 2x3 .......(ii)

∴\therefore∴ aaa ∝\propto∝ (distance)3

From equation (i), we get

v3/2=−1(t+5)3{v^{3/2}} = {{ - 1} \over {{{\left( {t + 5} \right)}^3}}}v3/2=(t+5)3−1​

Putting this in equation (ii), we get

aaa = -2v3/2{v^{3/2}}v3/2

∴\therefore∴ aaa ∝\propto∝ (velocity)3/2

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