Answer:
b) 6
Explanation:
Given
v(t)=3t²+6t
X(0) = 2
X(1) = ?
Knowing that
v(t)=3t²+6t = dX/dt
⇒ ∫dX = ∫(3t²+6t)dt
⇒ X - X₀ = t³ + 3t²
⇒ X(t) = X₀ + t³ + 3t²
If X(0) = 2
⇒ X(0) = X₀ + (0)³ + 3(0)² = 2
⇒ X₀ = 2
then we have
X(t) = t³ + 3t² + 2
when
t = 1
X(1) = (1)³ + 3(1)² + 2
X(1) = 6
After a time, t =1, the position of the particle has been 6. Thus, the correct option is b.
The velocity of the particle moving along the x-axis has been given by:
[tex]\rm v(t)\;=\;3t^2\;+\;6t[/tex]
The differentiation of v(t) in terms of x will be:
[tex]\rm \dfrac{dx}{dt}\;=\;3t^2\;+\;6t[/tex]
[tex]\rm \int dx\;=\;\int (3t^2\;+\;6t)\;dt[/tex]
differentiation with the limits if X be x and [tex]\rm x_0[/tex]:
x - [tex]\rm x_0[/tex] = [tex]\rm t^3\;+\;3t^2[/tex]
In the given question, the value of [tex]\rm x_0[/tex] = 2:
At time t = 0
x = [tex]\rm t^3\;+\;3t^2\;+\;x_0[/tex]
x = [tex]\rm (0)^3\;+\;3(0)^2\;+\;2[/tex]
x = 2.
To find the position of the particle at time =1, given, [tex]\rm x_0[/tex] = 2.
x = [tex]\rm (1)^3\;+\;3(1)^2\;+2[/tex]
x = 1 + 3 + 2
x = 6.
Thus, after time, t =1, the position of the particle has been 6. Thus, the correct option is b.
For more information about the position of the particle, refer to the link:
https://brainly.com/question/18328170
