Two cars start off to race with velocities `4 m/s and 2 m/s` and travel in straight line with uniform accelerations `1 msec^-2`and 2 msec − 2` respectively. If they reach the final point at the same instant, then the length of the path is.

Options

A. `30m`
B. `32m`
C. `20m`
D. `24m`

Answer

To solve the problem, we need to find the length of the path (L) that both cars travel, given their initial velocities and accelerations.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- For Car 1:
- Initial velocity (u₁) = 4 m/s
- Acceleration (a₁) = 1 m/s²
- For Car 2:
- Initial velocity (u₂) = 2 m/s
- Acceleration (a₂) = 2 m/s²

2. **Use the Second Equation of Motion:**
The equation for distance traveled under uniform acceleration is:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \(s\) is the distance traveled,
- \(u\) is the initial velocity,
- \(a\) is the acceleration,
- \(t\) is the time.

3. **Write the Equation for Car 1:**
For Car 1:
\[ L = u₁ t + \frac{1}{2} a₁ t^2 \]
Substituting the values:
\[ L = 4t + \frac{1}{2} \cdot 1 \cdot t^2 = 4t + \frac{1}{2} t^2 \]
This can be rewritten as:
\[ L = 4t + 0.5t^2 \quad \text{(Equation 1)} \]

4. **Write the Equation for Car 2:**
For Car 2:
\[ L = u₂ t + \frac{1}{2} a₂ t^2 \]
Substituting the values:
\[ L = 2t + \frac{1}{2} \cdot 2 \cdot t^2 = 2t + t^2 \]
This can be rewritten as:
\[ L = 2t + t^2 \quad \text{(Equation 2)} \]

5. **Set the Two Equations Equal:**
Since both cars travel the same distance (L) and reach the final point at the same time:
\[ 4t + 0.5t^2 = 2t + t^2 \]

6. **Rearrange the Equation:**
Rearranging gives:
\[ 4t + 0.5t^2 - 2t - t^2 = 0 \]
Simplifying:
\[ 2t - 0.5t^2 = 0 \]

7. **Factor the Equation:**
Factoring out \(t\):
\[ t(2 - 0.5t) = 0 \]
This gives us two solutions:
\[ t = 0 \quad \text{or} \quad 2 - 0.5t = 0 \]

8. **Solve for t:**
From \(2 - 0.5t = 0\):
\[ 0.5t = 2 \implies t = 4 \text{ seconds} \]

9. **Calculate the Length of the Path (L):**
Substitute \(t = 4\) seconds into either Equation 1 or Equation 2. Using Equation 2:
\[ L = 2(4) + (4)^2 = 8 + 16 = 24 \text{ meters} \]

### Final Answer:
The length of the path is \(L = 24 \text{ meters}\).

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