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

Options

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

Explanation

To solve the problem, we need to find the length of the path that both cars travel when they reach the final point at the same instant. Let's break this down step by step.

### Step 1: Define the variables
Let:
- Car 1 has an initial velocity \( u_1 = 4 \, \text{m/s} \) and acceleration \( a_1 = 1 \, \text{m/s}^2 \).
- Car 2 has an initial velocity \( u_2 = 2 \, \text{m/s} \) and acceleration \( a_2 = 2 \, \text{m/s}^2 \).
- Let \( t \) be the time taken for both cars to reach the final point.

### Step 2: Write the equations of motion for both cars
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
For Car 1:
\[ s = 4t + \frac{1}{2} \cdot 1 \cdot t^2 = 4t + \frac{1}{2} t^2 \]

For Car 2:
\[ s = 2t + \frac{1}{2} \cdot 2 \cdot t^2 = 2t + t^2 \]

### Step 3: Set the equations equal to each other
Since both cars travel the same distance \( s \) when they reach the final point at the same instant, we can set the two equations equal:
\[ 4t + \frac{1}{2} t^2 = 2t + t^2 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 4t + \frac{1}{2} t^2 - 2t - t^2 = 0 \]
\[ 2t - \frac{1}{2} t^2 = 0 \]

### Step 5: Factor out common terms
Factoring out \( t \):
\[ t \left( 2 - \frac{1}{2} t \right) = 0 \]
This gives us two solutions:
1. \( t = 0 \) (which we discard as it represents the starting point)
2. \( 2 - \frac{1}{2} t = 0 \) leading to:
\[ \frac{1}{2} t = 2 \implies t = 4 \, \text{s} \]

### Step 6: Calculate the length of the path
Now, we can substitute \( t = 4 \) seconds back into either equation to find the length of the path \( s \). Using Car 1's equation:
\[ s = 4(4) + \frac{1}{2} (4^2) \]
\[ s = 16 + \frac{1}{2} \cdot 16 = 16 + 8 = 24 \, \text{m} \]

### Final Answer
The length of the path is \( 24 \, \text{m} \).

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