Two particles $$ A $$ and $$ R $$ start to move at the same instant from a point $$ O $$. The particles move in the same direction along the same straight line. The acceleration of $$ A $$ at time $$ t $$ s after starting to move is $$ a \mathrm{~ms} \mathbf{s}^{-2} $$, where $$ a=0.05-0.0002t $$. i) Find $$ A^{ ext{'} } $$ s velocity when $$ t=200 $$ and when $$ t=500 $$. \ii) Find the distance between $$ A $$ and $$ B $$ when $$ t=500 $$.

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Two particles $$ A $$ and $$ R $$ start to move at the same instant from a point $$ O $$. The particles move in the same direction along the same straight line. The acceleration of $$ A $$ at time $$ t $$ s after starting to move is $$ a \mathrm{~ms} \mathbf{s}^{-2} $$, where $$ a=0.05-0.0002t $$. 	i) Find $$ A^{	ext{'} } $$ s velocity when $$ t=200 $$ and when $$ t=500 $$. \ii) Find the distance between $$ A $$ and $$ B $$ when $$ t=500 $$.

UpStudy AI · Accepted Answer

Let's analyze the motion of particle $$ A $$ and determine the required quantities step by step.

### Given:
- **Acceleration of $$ A $$** at time $$ t $$ seconds: 
  $$
  a(t) = 0.05 - 0.0002t \ \text{m/s}^2
  $$
- **Initial Conditions**:
  - Both particles start from rest at point $$ O $$.
  - $$ A $$ and $$ R $$ move in the same direction along the same straight line.

---

### Part i) Finding $$ A $$'s Velocity at $$ t = 200 $$ s and $$ t = 500 $$ s

To find the velocity of $$ A $$ at any time $$ t $$, we integrate the acceleration with respect to time.

$$
v(t) = \int a(t) \, dt + C
$$

Given that $$ A $$ starts from rest, the initial velocity $$ v(0) = 0 $$, so the constant of integration $$ C = 0 $$.

$$
v(t) = \int (0.05 - 0.0002t) \, dt = 0.05t - 0.0001t^2
$$

Now, calculate the velocity at the specified times:

1. **At $$ t = 200 $$ s**:
   $$
   v(200) = 0.05 \times 200 - 0.0001 \times (200)^2 = 10 - 4 = 6 \ \text{m/s}
   $$

2. **At $$ t = 500 $$ s**:
   $$
   v(500) = 0.05 \times 500 - 0.0001 \times (500)^2 = 25 - 25 = 0 \ \text{m/s}
   $$

**Summary for Part i:**
- $$ v_A(200 \ \text{s}) = 6 \ \text{m/s} $$
- $$ v_A(500 \ \text{s}) = 0 \ \text{m/s} $$

---

### Part ii) Finding the Distance Between $$ A $$ and $$ R $$ at $$ t = 500 $$ s

Assuming particle $$ R $$ remains stationary (since no information about $$ R $$'s motion is provided), the distance between $$ A $$ and $$ R $$ at time $$ t $$ is simply the displacement of $$ A $$ from the origin $$ O $$.

To find the displacement $$ s(t) $$ of $$ A $$, integrate the velocity:

$$
s(t) = \int v(t) \, dt + C
$$

Again, since $$ A $$ starts from rest at $$ O $$, the initial displacement $$ s(0) = 0 $$, so $$ C = 0 $$.

$$
s(t) = \int (0.05t - 0.0001t^2) \, dt = 0.025t^2 - \frac{0.0001}{3}t^3
$$

Calculate the displacement at $$ t = 500 $$ s:

$$
s(500) = 0.025 \times (500)^2 - \frac{0.0001}{3} \times (500)^3 = 0.025 \times 250{,}000 - \frac{0.0001}{3} \times 125{,}000{,}000
$$
$$
s(500) = 6{,}250 - 4{,}166.\overline{6} = 2{,}083.\overline{3} \ \text{meters}
$$

**Summary for Part ii:**
- The distance between $$ A $$ and $$ R $$ at $$ t = 500 $$ s is approximately $$ 2083.\overline{3} $$ meters.

---

### Final Answers:

**i)**
- At $$ t = 200 $$ s, velocity of $$ A $$ is 6 m/s.
- At $$ t = 500 $$ s, velocity of $$ A $$ is 0 m/s.

**ii)**
- At $$ t = 500 $$ s, the distance between $$ A $$ and $$ R $$ is approximately 2083 meters.
At $$ t = 200 $$ s, $$ A $$'s velocity is 6 m/s. At $$ t = 500 $$ s, $$ A $$'s velocity is 0 m/s. At $$ t = 500 $$ s, the distance between $$ A $$ and $$ R $$ is approximately 2083 meters.

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