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The motion of a particle is defined by the relation [tex]x = 2t^3 - 9t^2 + 12t + 10[/tex], where [tex]x[/tex] and [tex]t[/tex] are expressed in feet and seconds, respectively. Determine the time, position, and acceleration of the particle when [tex]v = 2.00 \, \text{ft/s}[/tex]. (Round the final answer to two decimal places.)

The time, position, and acceleration of the particle when [tex]v = 2.00 \, \text{ft/s}[/tex] are:

Answer :

Explanation:

The motion of a particle is defined by the relation as:

[tex]x=2t^3-9t^2+12t+10[/tex]........(1)

Differentiating equation (1) wrt t we get:

[tex]v=\dfrac{dx}{dt}[/tex]

[tex]v=\dfrac{d(2t^3-9t^2+12t+10)}{dt}[/tex]

[tex]v=6t^2-18t+12[/tex]............(2)

When v = 2 ft/s

[tex]2=6t^2-18t+12[/tex]

[tex]6t^2-18t+10=0[/tex]

t₁ = 2.26 s

and t₂ = 0.73 s

Put the value of t₁ in equation (1) as :

[tex]x=2(2.26)^3-9(2.26)^2+12(2.26)+10[/tex]

x₁ = 14.23 ft

Put the value of t₂ in equation (1) as :

[tex]x=2(0.73)^3-9(0.73)^2+12(0.73)+10[/tex]

x₁ = 14.74 ft

For acceleration differentiate equation (2) wrt t as :

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]a=\dfrac{d(6t^2-18t+12)}{dt}[/tex]

a = 12 t - 18.........(3)

Put t₁ and t₂ in equation 3 one by one as :

[tex]a_1=12(2.26)-18=9.12\ ft/s^2[/tex]

[tex]a_2=12(0.73)-18=-9.24\ ft/s^2[/tex]

Hence, this is the required solution.

The time taken by the particle is 2.26 s and the magnitude of acceleration and position of the particle at the obtained instant are [tex]9.12 \;\rm ft/s^{2}[/tex] and 14.23 ft respectively.

What is linear motion?

When an object is observed to move in a straight line, then the motion of the object is known as linear motion. It can be referenced as motion in one dimension.

Given data:

The position of the particle is, [tex]x = 2t^{3}-9t^{2}+12t+10[/tex].

The velocity of the particle is, [tex]v = 2.00 \;\rm ft/s[/tex].

We know that velocity can be obtained by differentiating the position. Then,

[tex]v =\dfrac{dx}{dt}\\\\v = \dfrac{d(2t^{3}-9t^{2}+12t+10)}{dt}\\\\v=6t^{2}-18t+12[/tex]

Now when v = 2.00 ft/s. The time is,

[tex]2 = 6t^{2}-18t+12\\\\6t^{2}-18t+10=0\\\\t = 2.26 \;\rm s[/tex]

Now, the position of the particle is,

[tex]x = 2t^{3}-9t^{2}+12t+10\\\\x = (2 \times 2.26^{3})-(9 \times 2.26^{2})+(12 \times 2.26)+10\\\\x =14.23\;\rm ft[/tex]

And the magnitude of the acceleration is calculated by differentiating the velocity as,

[tex]a = \dfrac{dv}{dt}\\\\a = \dfrac{d(6t^{2}-18t+12)}{dt}\\\\a=12t-18[/tex]

Substituting t = 2.26 s as,

[tex]a = 12(2.26)-18\\\\a=9.12 \;\rm ft/s^{2}[/tex]

Thus, we can conclude that the time taken by the particle is 2.26 s and the magnitude of acceleration and position of the particle at the obtained instant are [tex]9.12 \;\rm ft/s^{2}[/tex] and 14.23 ft respectively.

Learn more about the linear motion here:

https://brainly.com/question/848468

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