Respuesta :
Answer:
Explanation:
Given
[tex]v_x(t)=\alpha -\beta t^2[/tex]
[tex]\alpha =4\ m/s[/tex]
[tex]\beta =2\ m/s^3[/tex]
[tex]v_x(t)=4-2t^2[/tex]
[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]\int dx=\int \left ( 4-2t^2\right )dt[/tex]
[tex]x=4t-\frac{2}{3}t^3[/tex]
acceleration of object
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}[/tex]
[tex]a=-4t[/tex]
(b)For maximum positive displacement velocity must be zero at that instant
i.e.[tex]v=0[/tex]
[tex]4-2t^2=0[/tex]
[tex]t=\pm \sqrt{2}[/tex]
substitute the value of t
[tex]x=4\times \sqrt{2}-\frac{2}{3}\times 2\sqrt{2}[/tex]
[tex]x=3.77\ m[/tex]
The definitions of acceleration and velocity allow to find the results for the questions about the motion of the particle are:
A) the function of the acceleration is: a = -4t
and the position function is: x = 4 t - ⅔ t³
B) The maximum displacement is: x = 3.77 m
Given parameters
- The velocity of the body v = α-β t² with α = 4 m/s and β = 2 m/s²
To find
a) position and acceleration as a function of time,
b) maximum displacement,
The acceleration of defined as the change in velocity with time.
a = [tex]\frac{dv}{dt}[/tex]
Let's calculate.
a = - 2βt
a = - 2 2 t
a = -4 t
The speed is defined by the variation of the position with respect to time.
v = [tex]\frac{dx}{dt}[/tex]
dx = v dt
We integrate.
∫ dx = ∫ v dt
x - x₀ = ∫ (α - β t²)
x-x₀ = αt - βt³/ 3
we substitute.
x = 4 t - ⅔ t³
B) To find the maximum displacement we use the first derivative to be zero.
[tex]\frac{dx}{dt}[/tex] = 0
4 - 2t² = 0
t² = 2
t = √2 = 1.414 s
Let's find the position for this time.
x = 4 √2 - ⅔ (√2)³
x = 3.77 m
In conclusion using the definitions of acceleration and velocity we can find the result for the questions about the motion of the particle are:
A) the function of the acceleration is: a = -4t
and the position function is: x = 4 t - ⅔ t³
B) The maximum displacement is: x = 3.77 m
Learn more here: brainly.com/question/14683118
