The position of a particle is given by the function x=(4t3−6t2+12)m, where t is in s.
A.) at what time does the particle reach its minimum velocity
B.) what is (vx)min
C.) at what time is the acceleration zero

Respuesta :

Answer

given,

x = 4 t³ - 6 t² + 12

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

[tex]\dfrac{dx}{dt}=\dfrac{d}{dt}(4t^3-6t^2+12)[/tex]

[tex]v =12t^2-12t[/tex]

For minimum velocity calculation we have differentiate it and put it equal to zero.

[tex]\dfrac{dv}{dt} =\dfrac{d}{dt}12t^2-12t[/tex]

[tex]\dfrac{dv}{dt} =24t-12[/tex].........(1)

putting it equal to zero

24 t - 12 =0

t = 0.5 s

At t = 0.5 s velocity will be minimum.

b) minimum velocity

    v = 12t² -12 t

    v = 12 x 0.5² -12 x 0.5

    v = -3 m/s

c) derivative of velocity w.r.t. time is acceleration

from equation 1

     a = 24 t - 12

time at which acceleration will be zero

     0 = 24 t - 12

     t = 0.5 s

At t = 0.5 s acceleration will be zero.

Part A. The particle reaches its minimum velocity at 0.5 seconds.

Part B. The minimum velocity of the particle is -3 m/s.

Part C. The acceleration of the particle will be zero at the time t = 0.5 seconds.

How do you calculate the minimum velocity and acceleration?

Given that the position of a particle is given by the function x.

[tex]f(x) = 4t^2 -6t^2 +12[/tex]

The function of the velocity of a particle can be obtained by the time function.

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

[tex]v = \dfrac {d}{dt} ( 4t^3-6t^2 +12)[/tex]

[tex]v = 12t^2 -12 t[/tex]

The velocity function of the particle is [tex]v = 12t^2 - 12t[/tex].

Part A

The minimum velocity of the particle is obtained by the differentiation of velocity function with respect to the time and put it equal to zero.

[tex]\dfrac {dv}{dt} = \dfrac {d}{dt} (12t^2 - 12t) = 0[/tex]

[tex]\dfrac {dv}{dt} = 24 t-12 = 0[/tex]

[tex]t = 0.5\;\rm s[/tex]

Hence we can conclude that the particle reaches its minimum velocity at 0.5 seconds.

Part B

The velocity function is [tex]v = 12t^2 - 12t[/tex]. Substituting the value of t = 0.5 s to calculate the minimum velocity.

[tex]v = 12(0.5)^2 - 12(0.5)[/tex]

[tex]v = 3 - 6[/tex]

[tex]v = -3 \;\rm m/s[/tex]

The minimum velocity of the particle is -3 m/s.

Part C

The acceleration is defined as the change in the velocity with respect to time. Hence,

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

[tex]a = 24 t-12[/tex]

Substituting the value of a = 0, we get the time.

[tex]0 = 24t - 12[/tex]

[tex]t = 0.5 \;\rm s[/tex]

The acceleration of the particle will be zero at the time t = 0.5 seconds.

To know more about acceleration and velocity, follow the link given below.

https://brainly.com/question/2239252.