A scooter starts from rest have an acceleration of `1 ms^(-2)` while a car 150 m behind it starts from rest with an acceleration of `2ms^(-2)`. After how much time the car catches up with the scooter ?

Options

A. `sqrt(700)s`
B. `sqrt(300)s`
C. `sqrt(150)s`
D. None of the above

Explanation

To solve the problem step by step, we need to analyze the motion of both the scooter and the car.

### Step 1: Define the motion equations for both vehicles

1. **Scooter:**
- Initial velocity (u_scooter) = 0 m/s (starts from rest)
- Acceleration (a_scooter) = 1 m/s²
- Distance traveled by the scooter (x_scooter) after time t can be calculated using the equation of motion:
\[ x_{\text{scooter}} = u_{\text{scooter}} t + \frac{1}{2} a_{\text{scooter}} t^2 \]
- Substituting the values:
\[ x_{\text{scooter}} = 0 \cdot t + \frac{1}{2} \cdot 1 \cdot t^2 = \frac{1}{2} t^2 \]

2. **Car:**
- Initial velocity (u_car) = 0 m/s (starts from rest)
- Acceleration (a_car) = 2 m/s²
- The car starts 150 m behind the scooter, so the distance traveled by the car (x_car) after time t is:
\[ x_{\text{car}} = u_{\text{car}} t + \frac{1}{2} a_{\text{car}} t^2 \]
- Substituting the values:
\[ x_{\text{car}} = 0 \cdot t + \frac{1}{2} \cdot 2 \cdot t^2 = t^2 \]

### Step 2: Set up the equation for when the car catches up with the scooter

- The car catches up with the scooter when the distance traveled by the car equals the distance traveled by the scooter plus the initial distance between them (150 m). Therefore, we can write:
\[ x_{\text{car}} = x_{\text{scooter}} + 150 \]
- Substituting the expressions we derived:
\[ t^2 = \frac{1}{2} t^2 + 150 \]

### Step 3: Solve for time t

1. Rearranging the equation:
\[ t^2 - \frac{1}{2} t^2 = 150 \]
\[ \frac{1}{2} t^2 = 150 \]
\[ t^2 = 150 \times 2 = 300 \]
\[ t = \sqrt{300} \]

2. Simplifying:
\[ t = \sqrt{100 \times 3} = 10\sqrt{3} \text{ seconds} \]

### Final Answer:
The car catches up with the scooter after \( t = 10\sqrt{3} \) seconds.

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