Two cars X and Y start off to a race on a straight path with initial velocities of 8 m/s and 5 m/s respectively. Car X moves with uniform acceleration of `1 "m/s"^2` and car Y moves with uniform acceleration of `1.1 "m/s"^2`. If both the cars reach the winning post together find the length of the track.

Options

A. 1000 m
B. 2000m
C. 2500 m
D. 2280 m

Answer

To solve the problem, we need to find the length of the track such that both cars X and Y reach the winning post at the same time. We will use the equations of motion for uniformly accelerated motion.

### Step-by-Step Solution:

1. **Identify the known values:**
- For Car X:
- Initial velocity (u₁) = 8 m/s
- Acceleration (a₁) = 1 m/s²
- For Car Y:
- Initial velocity (u₂) = 5 m/s
- Acceleration (a₂) = 1.1 m/s²

2. **Use the second equation of motion:**
The distance (s) covered by an object under uniform acceleration can be calculated using the formula:
\[ s = ut + \frac{1}{2} a t^2 \]
Since both cars reach the winning post together, we can set their distances equal to each other.

3. **Write the equations for both cars:**
- For Car X:
\[ s = u₁ t + \frac{1}{2} a₁ t^2 \]
Substituting the values:
\[ s = 8t + \frac{1}{2} (1)t^2 = 8t + 0.5t^2 \]
- For Car Y:
\[ s = u₂ t + \frac{1}{2} a₂ t^2 \]
Substituting the values:
\[ s = 5t + \frac{1}{2} (1.1)t^2 = 5t + 0.55t^2 \]

4. **Set the distances equal to each other:**
\[ 8t + 0.5t^2 = 5t + 0.55t^2 \]

5. **Rearrange the equation:**
\[ 8t - 5t + 0.5t^2 - 0.55t^2 = 0 \]
Simplifying gives:
\[ 3t - 0.05t^2 = 0 \]
Factoring out t:
\[ t(3 - 0.05t) = 0 \]
This gives us two solutions: \( t = 0 \) (not valid) or \( 3 - 0.05t = 0 \).

6. **Solve for t:**
\[ 0.05t = 3 \implies t = \frac{3}{0.05} = 60 \text{ seconds} \]

7. **Calculate the length of the track (s):**
Using the equation for Car X:
\[ s = 8t + 0.5t^2 \]
Substituting \( t = 60 \):
\[ s = 8(60) + 0.5(60^2) = 480 + 0.5(3600) = 480 + 1800 = 2280 \text{ meters} \]

### Final Answer:
The length of the track is **2280 meters**.