The same retarding force is applied to stop a train. IF the speed is doubled then the distance will be

Options

A. the same
B. doubled
C. half
D. four times

Answer

To solve the problem, we can use the equations of motion. Let's break down the steps:

### Step 1: Understand the problem
We need to determine how the stopping distance of a train changes when its initial speed is doubled while the same retarding force is applied.

### Step 2: Use the equation of motion
We will use the third equation of motion:
\[ V^2 - U^2 = 2aS \]
Where:
- \( V \) = final velocity (0 when the train stops)
- \( U \) = initial velocity
- \( a \) = acceleration (negative for retarding force)
- \( S \) = distance traveled while stopping

### Step 3: Set up the equation for the first case
In the first case, let the initial speed be \( U \) and the final speed \( V = 0 \). The equation becomes:
\[ 0^2 - U^2 = 2(-a)S_1 \]
This simplifies to:
\[ -U^2 = -2aS_1 \]
Rearranging gives:
\[ S_1 = \frac{U^2}{2a} \]

### Step 4: Set up the equation for the second case
In the second case, the initial speed is doubled, so \( U' = 2U \). The final speed remains \( V = 0 \). The equation becomes:
\[ 0^2 - (2U)^2 = 2(-a)S_2 \]
This simplifies to:
\[ -4U^2 = -2aS_2 \]
Rearranging gives:
\[ S_2 = \frac{4U^2}{2a} = \frac{2U^2}{a} \]

### Step 5: Relate the two distances
Now we can relate \( S_2 \) to \( S_1 \):
\[ S_2 = 4 \left(\frac{U^2}{2a}\right) = 4S_1 \]
This means that if the speed is doubled, the stopping distance becomes four times the original distance.

### Conclusion
Thus, if the speed of the train is doubled, the distance required to stop the train will be four times the distance required at the original speed.

### Final Answer
If the speed is doubled, the distance will be **four times** the original distance.

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