A car accelerates from rest at constant rate for the first 10 s and covers a distance x. It covers a distance y in the next 10 s at the same acceleration. Which of the following is true?

Options

A. `x=3y `
B. `y = 3x`
C. `x=y`
D. `y=2x`

Answer

To solve the problem step-by-step, we will analyze the motion of the car during the two intervals of time given in the question.

### Step 1: Analyze the first 10 seconds
The car starts from rest, which means the initial velocity \( u = 0 \). We are given that the car accelerates at a constant rate \( \alpha \) for the first 10 seconds.

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ x = 0 \cdot 10 + \frac{1}{2} \alpha (10)^2 \]
This simplifies to:
\[ x = \frac{1}{2} \alpha \cdot 100 = 50\alpha \]
So, the distance covered in the first 10 seconds is:
\[ x = 50\alpha \quad \text{(Equation 1)} \]

### Step 2: Analyze the next 10 seconds
In the next 10 seconds (from 10 seconds to 20 seconds), the car continues to accelerate at the same rate \( \alpha \). The total time now is 20 seconds.

Using the same equation of motion for the total distance covered in 20 seconds:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ x + y = 0 \cdot 20 + \frac{1}{2} \alpha (20)^2 \]
This simplifies to:
\[ x + y = \frac{1}{2} \alpha \cdot 400 = 200\alpha \]
So, we have:
\[ x + y = 200\alpha \quad \text{(Equation 2)} \]

### Step 3: Relate the two equations
Now we have two equations:
1. \( x = 50\alpha \) (from the first 10 seconds)
2. \( x + y = 200\alpha \) (from the total 20 seconds)

We can substitute Equation 1 into Equation 2:
\[ 50\alpha + y = 200\alpha \]
Now, solving for \( y \):
\[ y = 200\alpha - 50\alpha = 150\alpha \quad \text{(Equation 3)} \]

### Step 4: Find the relationship between x and y
From Equation 1 and Equation 3, we have:
- \( x = 50\alpha \)
- \( y = 150\alpha \)

Now, we can find the ratio of \( y \) to \( x \):
\[ y = 3x \quad \text{(since } 150\alpha = 3 \cdot 50\alpha\text{)} \]

### Conclusion
Thus, the relationship between the distances covered in the two intervals is:
\[ y = 3x \]
This means that the correct statement is that the distance \( y \) covered in the second 10 seconds is three times the distance \( x \) covered in the first 10 seconds.