Simple but accelerating

Classical Mechanics Level 2

Velocity of a particle varies with its displacement as v 2 v^2 =9- x 2 x^2 m/s find the magnitude of maximum acceleration of the particle.(in SI units).

3 None of the above 2 4 1

Adarsh Adi by Adarsh Adi , 19, India

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2 solutions

Steven Chase
Feb 21, 2018

Implicit differentiation

v 2 = 9 − x 2 2 v a = − 2 x v a = − x v^2 = 9 - x^2 \\ 2 \, v \, a = -2 \, x \, v \\ a = -x

The square of the velocity cannot be negative. Therefore, the range on x x is [ − 3 , 3 ] [-3,3] . The maximum acceleration is therefore 3

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Adarsh Adi
Feb 21, 2018

Comparing the given equation with standard velocity- displacement equation of simple harmonic motion , i.e, ( v / w ) 2 (v/w)^2 = A 2 A^2 - x 2 x^2 ,we get w=1rad/sec and A=3m. The magnitude of maximum acceleration of the particle in SHM= w 2 w^2 A= 1 2 1^2 (3)m/ s e c 2 sec^2 = 3m/ s e c 2 sec^2

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