Let's break down the problem into its parts. We know the particle's velocity, acceleration components, and initial conditions, and we need to calculate the position and velocity at $t = 4.0 \, \text{s}$.

Given Data:

• Initial position: $x_0 = 0 \, \text{m}, \, y_0 = 0 \, \text{m}$
• Initial velocities: $v_{x0} = 0 \, \text{m/s}, \, v_{y0} = 6.8 \, \text{m/s}$
• Accelerations: $a_x = -3.2 \, \text{m/s}^2, \, a_y = 0 \, \text{m/s}^2$
• Time: $t = 4.0 \, \text{s}$

We will solve the problem using the kinematic equations of motion.

Part (a): Find the $x$- and $y$-positions of the particle at $t = 4.0 \, \text{s}$.

The general equations for position under constant acceleration are:

$x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2$ $y = y_0 + v_{y0} t + \frac{1}{2} a_y t^2$

x-position:

$x = 0 + (0)(4.0) + \frac{1}{2} (-3.2)(4.0)^2$ $x = \frac{1}{2} (-3.2)(16) = -25.6 \, \text{m}$

y-position:

Since $a_y = 0$, the y-component of motion is uniform, so: $y = 0 + (6.8)(4.0) + \frac{1}{2} (0)(4.0)^2$ $y = 27.2 \, \text{m}$

So, at $t = 4.0 \, \text{s}$: $x = -25.6 \, \text{m}, \quad y = 27.2 \, \text{m}$

Part (b): Find the velocity components at $t = 4.0 \, \text{s}$.

The velocity equations under constant acceleration are:

$v_x = v_{x0} + a_x t$ $v_y = v_{y0} + a_y t$

x-component of velocity:

$v_x = 0 + (-3.2)(4.0) = -12.8 \, \text{m/s}$

y-component of velocity:

Since $a_y = 0$, the y-component remains constant: $v_y = 6.8 + (0)(4.0) = 6.8 \, \text{m/s}$

So, the velocity components at $t = 4.0 \, \text{s}$ are: $v_x = -12.8 \, \text{m/s}, \quad v_y = 6.8 \, \text{m/s}$

Part (c): How does the speed of the particle change from $t = 0$ to $t = 4.0 \, \text{s}$?

The speed of the particle is the magnitude of the velocity vector:

At $t = 0$: $v(0) = \sqrt{v_{x0}^2 + v_{y0}^2} = \sqrt{0^2 + 6.8^2} = 6.8 \, \text{m/s}$

At $t = 4.0$: $v(4.0) = \sqrt{v_x^2 + v_y^2} = \sqrt{(-12.8)^2 + (6.8)^2}$ $v(4.0) = \sqrt{163.84 + 46.24} = \sqrt{210.08} \approx 14.49 \, \text{m/s}$

Thus, the speed of the particle increases from $6.8 \, \text{m/s}$ at $t = 0$ to approximately $14.49 \, \text{m/s}$ at $t = 4.0 \, \text{s}$.

Summary:

• (a) $x = -25.6 \, \text{m}, \quad y = 27.2 \, \text{m}$
• (b) $v_x = -12.8 \, \text{m/s}, \quad v_y = 6.8 \, \text{m/s}$
• (c) Speed increases from $6.8 \, \text{m/s}$ to approximately $14.49 \, \text{m/s}$.

Would you like more details or further clarifications?

Here are five related questions:

1. How would the results change if the acceleration in the y-direction were non-zero?
2. What would happen to the particle's path if the x-component of velocity were initially non-zero?
3. How can we determine the total displacement of the particle after 4.0 seconds?
4. What is the angle of the velocity vector at $t = 4.0 \, \text{s}$?
5. How would the motion differ if the accelerations were reversed in sign?

Tip: When calculating motion under constant acceleration, always check if any components of velocity or acceleration are zero to simplify the equations.