The motion of a particle is defined by the relation x = 2t3 – 9t2 +12t +10, where x and t are expressed in feet and seconds, respectively. determine the time, position, and acceleration of the particle when v = 2.00 ft/s. (round the final answer to two decimal places.) the time, the position, and the acceleration of the particle when v = 2.00 are:

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 and 14.23 ft respectively.

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, .

The velocity of the particle is, .

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

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

Now, the position of the particle is,

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

Substituting t = 2.26 s as,

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 and 14.23 ft respectively.

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Explanation:

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

........(1)

Differentiating equation (1) wrt t we get:

............(2)

When v = 2 ft/s

t₁ = 2.26 s

and t₂ = 0.73 s

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

x₁ = 14.23 ft

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

x₁ = 14.74 ft

For acceleration differentiate equation (2) wrt t as :

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

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

Hence, this is the required solution.

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