Question

The velocity of a particle moving along the x–axis is given by v = x3 – 6x2 + 12 where v is in m/s and x is in m. Acceleration of the particle when it is passing through the point x = 4 m will be​

Answer 1

Answer:

v=x

2

+x

⇒

dx

dv

=2x+1

∴a=

dx

vdv

=(x

2

+x)(2x+1)

∴atr=2

a=(4+2)(4+1)

=30m/s

2

∴ Option D is correct answer

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Answer 2

Given:

The velocity of the particle moving along the x-axis, v = x³ - 6x² + 12

To Find:

Acceleration of the particle when it is passing through the point x = 4m

Solution;

v = x³ - 6x² + 12

Differentiating with respect to x,

we get the acceleration of the particle

i.e. a = 3x² - 12x + 0

a (at x = 4m) = 3(4)² - 12(4)

a = 48 - 48 = 0

Hence, the correct answer is 0 m/s²

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