Question

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:

Answer 1

Explanation:

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

$x=2t^3-9t^2+12t+10$........(1)

Differentiating equation (1) wrt t we get:

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

$v=\dfrac{d(2t^3-9t^2+12t+10)}{dt}$

$v=6t^2-18t+12$............(2)

When v = 2 ft/s

$2=6t^2-18t+12$

$6t^2-18t+10=0$

t₁ = 2.26 s

and t₂ = 0.73 s

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

$x=2(2.26)^3-9(2.26)^2+12(2.26)+10$

x₁ = 14.23 ft

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

$x=2(0.73)^3-9(0.73)^2+12(0.73)+10$

x₁ = 14.74 ft

For acceleration differentiate equation (2) wrt t as :

$a=\dfrac{dv}{dt}$

$a=\dfrac{d(6t^2-18t+12)}{dt}$

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

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

$a_1=12(2.26)-18=9.12\ ft/s^2$

$a_2=12(0.73)-18=-9.24\ ft/s^2$

Hence, this is the required solution.