contestada

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position is x0 = 1.1 m at t0 = 0 s . part a at 2.0 s , what is the particle's position?

Respuesta :

Cricetus

The position of the object at time t =2.0 s is 6.4 m.

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

[tex]v_x= \frac{dx}{dt}[/tex]

Write an equation for x.

[tex]dx=v_xdt\\ x=\int v_xdt[/tex]

Substitute the equation for vₓ =2t² in the integral.

[tex]x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C[/tex]

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

[tex]x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m[/tex]

Substitute 2.0s for t.

[tex]x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m[/tex]

The position of the particle at t =2.0 s is 6.4m




whitneytr12

The particle's position along the x-axis when t = 2.0 s is 6.4 m. It was calculated knowing that the velocity is equal to [tex]v_{x} = 2t^{2} [/tex] and x₀ = 1.1 m when t₀ = 0 s.  

To find the particle's position we need to remember that:

[tex] v = \frac{dx}{dt} [/tex]  (1)    

Where:

x: is the position  

t: is the time

The distance is then:

[tex] dx = vdt [/tex]

We know that x₀ = 1.1 m when t₀ = 0s and since need to find the position when t = 2.0s, we have:

[tex] \int ^{x}_{1.1} dx = \int ^{2}_{0} vdt [/tex]  

Entering [tex]v = 2t^{2}[/tex] into the above equation:

[tex] \int ^{x}_{1.1} dx = \int ^{2}_{0} 2t^{2}dt [/tex]  

[tex] x|_{1.1}^{x} = \frac{2}{3}t^{3}|_{0}^{2} [/tex]

[tex] x - 1.1 = \frac{2}{3}(2^{3} - 0^{3}) [/tex]

[tex] x = 6.4 [/tex]

Therefore, the particle's position when t = 2.0 s is 6.4 m.

Find more here:

  • https://brainly.com/question/862972?referrer=searchResults
  • https://brainly.com/question/1584190?referrer=searchResults

I hope it helps you!

Ver imagen whitneytr12

Otras preguntas