Initially a car is moving with speed of `20 ms^(-1)` and covers a distance of 100 m before it comes to rest. On making the initial speed of car ` 40 ms^(-1)` keeping the acceleration same, the distance travelled by car before it comes to rest is

Options

A. 200 m
B. 300 m
C. 400 m
D. 800 m

Answer

To solve the problem step by step, we will use the equations of motion.

### Step 1: Identify the known values for the first case
- Initial speed (U1) = 20 m/s
- Final speed (V1) = 0 m/s (the car comes to rest)
- Distance (S1) = 100 m
- Acceleration (A1) = ?

### Step 2: Use the third equation of motion to find acceleration (A1)
The third equation of motion states:
\[ V^2 = U^2 + 2AS \]

Rearranging for acceleration (A):
\[ A = \frac{V^2 - U^2}{2S} \]

Substituting the known values:
\[ A1 = \frac{0^2 - (20)^2}{2 \times 100} \]
\[ A1 = \frac{0 - 400}{200} \]
\[ A1 = -2 \, \text{m/s}^2 \]

### Step 3: Identify the known values for the second case
- Initial speed (U2) = 40 m/s
- Final speed (V2) = 0 m/s (the car comes to rest)
- Acceleration (A2) = A1 = -2 m/s² (same as the first case)
- Distance (S2) = ?

### Step 4: Use the third equation of motion to find the distance (S2) for the second case
Using the same equation:
\[ V^2 = U^2 + 2AS \]

Rearranging for distance (S):
\[ S = \frac{V^2 - U^2}{2A} \]

Substituting the known values:
\[ S2 = \frac{0^2 - (40)^2}{2 \times (-2)} \]
\[ S2 = \frac{0 - 1600}{-4} \]
\[ S2 = \frac{-1600}{-4} \]
\[ S2 = 400 \, \text{m} \]

### Conclusion
The distance traveled by the car before it comes to rest when the initial speed is 40 m/s is **400 meters**.

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