Respuesta :
Answer:
The acceleration of the particle when it achieves its maximum displacement in the positive x direction is [tex]a_x=-57.9\ m/s^2[/tex].
Explanation:
The velocity of a particle moving along the x axis is given by :
[tex]v_x=at-bt^3[/tex]..........(1)
We need to find the acceleration of the particle when it achieves its maximum displacement in the positive x direction.
Put [tex]v_x=0[/tex]
[tex]at-bt^3=0[/tex]
[tex]29t-1.4t^3=0[/tex]
t = 4.55 seconds
Differentiating equation (1) we get :
[tex]a=\dfrac{dv_x}{dt}=\dfrac{d}{dt}(at-bt^3)[/tex]
[tex]a_x=a-3bt^2[/tex]
[tex]a_x=29-3\times 1.4\times (4.55)^2[/tex]
[tex]a_x=-57.9\ m/s^2[/tex]
So, the acceleration of the particle when it achieves its maximum displacement in the positive x direction is [tex]a_x=-57.9\ m/s^2[/tex]. Hence, this is the required solution.
Answer:
The acceleration of the particle is -57.94 m/s²
Explanation:
Given that,
The value of a = 29 m/s²
The value of b = 1.4 m/s⁴
The velocity of the particle is
[tex]v_{x}=at-bt^3[/tex]
We need to calculate the time
For maximum displacement,
At v = 0 in equation (I)
[tex]0=at-bt^3[/tex]
[tex]at=bt^3[/tex]
[tex]t^2=\dfrac{a}{b}[/tex]
Put the value into the formula
[tex]t^2=\dfrac{29}{1.4}\ s[/tex]
[tex]t^2=20.7\ sec[/tex]
We need to calculate the acceleration of the particle
On differential of velocity equation
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]a=\dfrac{d}{dt}(at-bt^3)[/tex]
[tex]a=a-3bt^2[/tex]
put the value into the formula
[tex]a=29-3\times1.4\times20.7[/tex]
[tex]a=57.94\ m/s^2[/tex]
Hence, The acceleration of the particle is -57.94 m/s²
