To solve this problem, we need to determine the particle's position, velocity, and acceleration at $t = 1\text{s}$.

1. Velocity at $t = 1\text{s}$

Given:

• Initial velocity at $t = 0.5\text{s}$: $v(0.5) = 5$ units per second.
• Acceleration: $a = -2$ units per second squared.
• Time step: $\Delta t = 0.5\text{s}$.

The velocity at $t = 1\text{s}$ can be found using the equation: $v(1) = v(0.5) + a \cdot \Delta t$ Substituting the values: $v(1) = 5 + (-2) \cdot 0.5 = 5 - 1 = 4 \text{ units per second}$

2. Position at $t = 1\text{s}$

Given:

• Initial position at $t = 0.5\text{s}$: $x(0.5) = 3$ units.
• Initial velocity at $t = 0.5\text{s}$: $v(0.5) = 5$ units per second.
• Acceleration: $a = -2$ units per second squared.
• Time step: $\Delta t = 0.5\text{s}$.

The position at $t = 1\text{s}$ can be found using the equation: $x(1) = x(0.5) + v(0.5) \cdot \Delta t + \frac{1}{2} a \cdot (\Delta t)^2$ Substituting the values: $x(1) = 3 + 5 \cdot 0.5 + \frac{1}{2} \cdot (-2) \cdot (0.5)^2$ $x(1) = 3 + 2.5 - 0.25 = 5.25 \text{ units}$

3. Acceleration at $t = 1\text{s}$

The acceleration is constant, so it remains: $a(1) = -2 \text{ units per second squared}$

Summary of Results

• Position at $t = 1\text{s}$: $x(1) = 5.25$ units.
• Velocity at $t = 1\text{s}$: $v(1) = 4$ units per second.
• Acceleration at $t = 1\text{s}$: $a(1) = -2$ units per second squared.

Would you like more details on any of these calculations or have any questions?

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Here are 5 related questions you might explore next:

1. How would the position change if the acceleration were positive instead of negative?
2. What would be the particle's position at $t = 1.5\text{s}$?
3. How does the acceleration affect the velocity over time?
4. What would be the final velocity if the particle continued to decelerate for another second?
5. How does the time step size affect the accuracy of the simulation?

Tip: Remember, when dealing with constant acceleration, you can always use kinematic equations to relate position, velocity, and time.