Question

A particle moving along the x -axis has a position given by x = ( 24t − 2 t^3 ) m , where 't' is measured in s . What is the magnitude of the acceleration of the particle at the instant when its velocity is zero?

Answer 1

As per the question , the displacement function is given as :

$x = 24t - 2{t}^{3}$

We can differentiate this function to get the instantaneous velocity function :

$v = \dfrac{dx}{dt}$

$= > v = 24- 6{t}^{2}$

We need to find the time at which velocity was zero :

$= > v = 24- 6{t}^{2} = 0$

$= > 6 {t}^{2} = 24$

$= > {t}^{2} = 4$

$= > t = 2 \: sec$

Now again differentiating the velocity function, we get instantaneous acceleration function :

$a = \dfrac{dv}{dt} = - 12t$

Putting value of t = 2 sec :

$= > a = - 12t$

$= > a = - 12 \times 2$

$= > a = - 24 \: m {s}^{ - 2}$

So final answer is :

Acceleration is 24 m/s² when velocity is zero.

Answer 2

Given :

• x = 24t - 2t³

To Find :

• Magnitude of acceleration when velocity is 0 m/s

Solution :

We are given that,

$\longrightarrow \sf{x \: = \: 24t \: - \: 2t^3}$

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$\underbrace{\sf{Velocity}}$

$\implies \sf{v \: = \: \dfrac{dx}{dt}} \\ \\ \implies \sf{v \: = \: \dfrac{d(24t \: - \: 2t^3)}{dt}} \\ \\ \implies \sf{v \: = \: 24 \: - \: 2(3)t^2} \\ \\ \implies \sf{v \: = \: 24 \: - \: 6t^2 \: ms^{-1}}$

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$\underbrace{\sf{Time \: at \: v \: = \: 0 \: ms^{-1}}}$

Now, velocity is 0 m/s

$\implies \sf{0 \: = \: 24 \: - \: 6t^2} \\ \\ \implies \sf{-6t^2 \: = \: -24} \\ \\ \implies \sf{6t^2 \: = \: 24} \\ \\ \implies \sf{t^2 \: = \: \dfrac{24}{6}} \\ \\ \implies \sf{t^2 \: = \: 4} \\ \\ \implies \sf{t \: = \: \sqrt{4}} \\ \\ \implies \sf{t \: = \: 2 \: s}$

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$\underbrace {\sf{Acceleration \: when \: v \: = \: 0 \: ms^{-1}}}$

$\implies \sf{a \: = \: \dfrac{dv}{dt}} \\ \\ \implies \sf{a \: = \: \dfrac{d(24 \: - \: 6t^2)}{dt}} \\ \\ \implies \sf{a \: = \: 0 \: - \: 6(2)t} \\ \\ \implies \sf{a \: = \: -12t}$

Put t = 2

$\implies \sf{a \: = \: -12(2)} \\ \\ \implies \sf{a \: = \: -24}$

$\therefore$ Acceleration is -24 m/s²