A car moving with a velocity of `20ms^(-1)` is stopped in a distance of 40m. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is

Options

A. 640m
B. 320m
C. 160m
D. 1280m

Explanation

To solve the problem step by step, we will use the equations of motion and the relationship between initial velocity, distance, and acceleration (retardation in this case).

### Step 1: Understand the given data
- Initial velocity of the car, \( u_1 = 20 \, \text{m/s} \)
- Distance traveled before stopping, \( s_1 = 40 \, \text{m} \)
- Final velocity when the car stops, \( v = 0 \, \text{m/s} \)

### Step 2: Use the equation of motion
We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
Since the car is stopping, the final velocity \( v = 0 \). Therefore, we can rewrite the equation as:
\[ 0 = u^2 + 2as \]
This can be rearranged to:
\[ u^2 = -2as \]
Since \( a \) is negative (retardation), we can write:
\[ u^2 = 2|a|s \]
Where \( |a| \) is the magnitude of retardation.

### Step 3: Calculate retardation for the first case
For the first case:
\[ u_1^2 = 2|a|s_1 \]
Substituting the known values:
\[ (20)^2 = 2|a|(40) \]
\[ 400 = 80|a| \]
\[ |a| = \frac{400}{80} = 5 \, \text{m/s}^2 \]

### Step 4: Determine the new initial velocity
Now, if the car is traveling at double the velocity:
\[ u_2 = 2 \times 20 = 40 \, \text{m/s} \]

### Step 5: Calculate the distance for the second case
Using the same equation of motion for the second case:
\[ u_2^2 = 2|a|s_2 \]
Substituting the known values:
\[ (40)^2 = 2(5)s_2 \]
\[ 1600 = 10s_2 \]
\[ s_2 = \frac{1600}{10} = 160 \, \text{m} \]

### Final Answer
The distance traveled by the car when traveling at double the velocity is \( s_2 = 160 \, \text{m} \).

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