Question

At time t - 0, a particle is at rest in the x-y plane at the coordinates (xo,yo) (4, 0) in. If the particle is then subjected to the acceleration components ax =0.38 -0.23t in. /sec2 and ay = 0.14t - 0.03t^2 in. /sec2, determine the coordinates of the particle position when t - 6 sec.

Answer 1

The coordinates of the particle position at t = 6 sec sec are: (x, y) ≈ (10.84, -27.72) in.

Step-by-step explanation:

Initial position of the particle: (x₀, y₀) = (4, 0) in

Acceleration components: ax = 0.38 - 0.23t in./sec² and ay = 0.14t - 0.03t² in./sec²

Time, t = 6 sec

The position of the particle at time t can be determined using the equations of motion. The equations of motion for the x and y coordinates are given by:

`x = x_0 + v_0xt + (1)/(2)axt^2\\\y = y_0 + v_0yt + (1)/(2)ayt^2\`

Where:

x₀ and y₀ are the initial positions,

v₀x and v₀y are the initial velocities,

ax and ay are the accelerations.

First, we need to find the velocity components v₀x and v₀y at time t = 6 sec.

The velocity components can be found using the following equations:

`v_x = v_0x + a_xt\\v_y = v_0y + a_yt`

Given that the particle is initially at rest, v₀x = v₀y = 0.

Now, let's calculate the velocity components at time t = 6 sec:

`v_x = 0 + \int_0^6 (0.38 - 0.23t) dt\\\\v_y = 0 + \int_0^6 (0.14t - 0.03t^2) dt`

Integrating with respect to time:

`v_x = \left[0.38t - (0.23t^2)/(2)\right]_0^6\\\\v_y = \left[0.14(t^2)/(2) - 0.03(t^3)/(3)\right]_0^6`

Solving for v_x and v_y:

`v_x = 0.38(6) - (0.23(6)^2)/(2) - (0)\\\\v_y = 0.14\left((6^2)/(2)\right) - 0.03\left((6^3)/(3)\right) - (0)`

`\[v_x = 2.28 - 4.14\]\\v_y = 1.68 - 3.24\]`

`\[v_x = -1.86\text{ in/sec}\]\\v_y = -1.56\text{ in/sec}\]`

Now that we have the velocity components, we can calculate the final position of the particle at time t = 6 sec.

Using the equations of motion:

`\[x = x_0 + v_0xt + (1)/(2)axt^2\]\\y = y_0 + v_0yt + (1)/(2)ayt^2\]`

Plugging in the values:

`\[x = 4 + (0)(6) + (1)/(2)(0.38)(6)^2\]\\`

`\[y = 0 + (0)(6) + (1)/(2)(0.14)(6)^2 - (1)/(2)(0.03)(6)^3\]`

Solving for x and y:

`\[x = 4 + 6.84\]\\y = 4.68 - 32.4\]`

Therefore, at t=6 sec, the particle's position is approximately (10.84, -27.72) inches.