contestada

At t = 0, a particle starts at rest and moves along a line in such a way that, at time t, its acceleration is 24t 2 ft / s2. through how many feet does the particle move during the first 2 seconds?

Respuesta :

skyluke89
The acceleration of the particle at time t is:
[tex]a(t)=24t^2 ft/s^2[/tex]
The velocity of the particle at time t is given by the integral of the acceleration a(t):
[tex]v(t)= \int a(t) \, dt = \int (24 t^2) dt=24 \frac{t^3}{3}=8t^3 ft/s [/tex]
and the position of the particle at time t is given by the integral of the velocity v(t):
[tex]x(t)=\int v(t) = \int (8t^3)=8 \frac{t^4}{4}=2t^4 ft [/tex]

Assuming the particle starts from position x(0)=0 at t=0, the distance the particle covers in the first t=2 seconds can be found by substituting t=2 s in the equation of x(t):
[tex]x(2 s)=2 t^4 = 2 (2s)^4=32 ft[/tex]
Lanuel

The distance in feet covered by this particle during the first 2 seconds is 32 feet.

Given the following data:

  • Initial time = 0 seconds
  • Acceleration = [tex]24t^2 \;ft/s^2[/tex]

To calculate the distance in feet covered by this particle during the first 2 seconds:

First of all, we would determine the velocity at which the particle is moving by integrating the value of its acceleration as follows;

[tex]V(t) = \int\limits {(a)} \, dt \\\\V(t) = \int {(24t^2)} \, dt\\\\V(t) = 24 \frac{t^3}{3} \\\\V(t) = 8t^3\; ft/s[/tex]

Next, we would calculate the position of this particle by integrating the velocity with respect to time;

[tex]x(t) = \int\limits {(V)} \, dt \\\\x(t) = \int {(8t^3)} \, dt\\\\V(t) = 8 \frac{t^4}{4} \\\\x(t) = 2t^4\; ft[/tex]

Now, we can calculate the distance covered by this particle during the first 2 seconds:

[tex]Distance = 2\times 2^4\\\\Distance = 2\times 16[/tex]

Distance = 32 feet.

Read more on distance here: https://brainly.com/question/10545161

Otras preguntas