Question

a particles moves According to the law x = t³-6t²+9t+5 the displacement of the particle at the time when its acceleration is zero ...
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Answer 1

Question :-

• A particles moves according to the law x = t³-6t²+9t+5, then find the displacement of the particle at the time when its acceleration is zero.

Answer:-

Given :-

• A particle moves according as x = t³-6t²+9t+5.
• Accelaration of the particle is 0.

To find :

• Displacement when Accelaration is 0.

Formula used -

$\bf \:Velocity \: (v) = \dfrac{dx}{dt}$

$\bf \:Accelaration \: (a) = \dfrac{dv}{dt}$

$\bf \:\dfrac{d}{dx} {x}^{n} = n {x}^{n - 1}$

$\bf \:\dfrac{d}{dx}k = 0.$

Solution:-

Consider, = t³-6t²+9t+5

Differentiate w. r. t. t, we get,

$\bf \:\dfrac{dx}{dt} = \dfrac{d}{dt} ( t³-6t²+9t+5)$

$\bf\implies \:\dfrac{dx}{dt} = \dfrac{d}{dt} {t}^{3} - 6\dfrac{d}{dt} {t}^{2} + 9\dfrac{d}{dt}t + \dfrac{d}{dt}5$

$\bf\implies \:v = {3t}^{2} - 12t + 9$

Differentiate w. r. t. t, we get

$\bf \:\dfrac{d}{dt}v = \dfrac{d}{dt}( {3t}^{2} - 12t + 9)$

$\bf\implies \:a = 3\dfrac{d}{dt} {t}^{2} - 12\dfrac{d}{dt}t + \dfrac{d}{dt}9$

$\bf\implies \:a = 6t - 12$

As accelaration = 0

$\bf\implies \:6t - 12 = 0$

$\bf\implies \:t = 2 units$

So, displacement when Accelaration is zero is

$\bf\implies \:x(2) = {2}^{3} - 6 {(2)}^{2} + 9 \times 2 + 5$

$\bf\implies \:x(2) = 8 - 24 + 18 + 5 = 7 \: units$

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