A car travelling with a velocity of 80 km/h slowed down to 44 km/h in 15 s. The retardation is

Options

A. `0.67m//s^(2)`
B. `1m//s^(2)`
C. `1.25m//s^(2)`
D. `1.5m//s^(2)`

Answer

To solve the problem of finding the retardation of a car that slows down from 80 km/h to 44 km/h in 15 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Convert velocities from km/h to m/s:**
- Initial velocity (u) = 80 km/h
- Final velocity (V) = 44 km/h
- To convert km/h to m/s, we use the conversion factor: \(1 \text{ km/h} = \frac{5}{18} \text{ m/s}\).
- Therefore,
\[ u = 80 \times \frac{5}{18} = \frac{400}{18} \approx 22.22 \text{ m/s} \]
\[ V = 44 \times \frac{5}{18} = \frac{220}{18} \approx 12.22 \text{ m/s} \]

2. **Use the formula for acceleration:**
- The formula relating initial velocity (u), final velocity (V), acceleration (a), and time (t) is:
\[ V = u + a \cdot t \]
- Rearranging the formula to solve for acceleration (a):
\[ a = \frac{V - u}{t} \]

3. **Substitute the known values into the equation:**
- We know:
- \(V = 12.22 \text{ m/s}\)
- \(u = 22.22 \text{ m/s}\)
- \(t = 15 \text{ s}\)
- Plugging in these values:
\[ a = \frac{12.22 - 22.22}{15} \]
\[ a = \frac{-10}{15} = -0.67 \text{ m/s}^2 \]

4. **Interpret the result:**
- The negative sign indicates that this is a retardation (deceleration).
- Therefore, the magnitude of the retardation is:
\[ \text{Retardation} = 0.67 \text{ m/s}^2 \]

### Final Answer:
The retardation of the car is \(0.67 \text{ m/s}^2\).

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