A car left 3 minutes early than the scheduled time and in order to reach the destination 126km away in time, it has to slow its speed by 6km/h from the usual. What is the usual speed (in km/hr) of the car?

Options

A. A) 56
B. B) 63
C. C) `94.5`
D. D) 126

Answer

To solve the problem step by step, we can use the relationship between distance, speed, and time. Here’s how we can approach it:

### Step 1: Understand the problem
The car leaves 3 minutes early and has to reduce its speed by 6 km/h to reach the destination on time. We need to find the usual speed of the car.

### Step 2: Convert time into hours
Since the time difference is given in minutes, we need to convert it into hours for our calculations.
- 3 minutes = 3/60 hours = 1/20 hours.

### Step 3: Set up the equations
Let the usual speed of the car be \( V \) km/h. If the car slows down by 6 km/h, its new speed becomes \( V - 6 \) km/h.

### Step 4: Calculate time taken at both speeds
1. **Time taken at usual speed**:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{126}{V} \text{ hours} \]

2. **Time taken at reduced speed**:
\[ \text{Time} = \frac{126}{V - 6} \text{ hours} \]

### Step 5: Set up the equation based on the time difference
Since the car leaves 3 minutes early, the time difference between the two speeds is 1/20 hours. Therefore, we can set up the equation:
\[ \frac{126}{V - 6} - \frac{126}{V} = \frac{1}{20} \]

### Step 6: Solve the equation
Multiply through by \( V(V - 6) \) to eliminate the denominators:
\[ 126V - 126(V - 6) = \frac{V(V - 6)}{20} \]

This simplifies to:
\[ 126V - 126V + 756 = \frac{V^2 - 6V}{20} \]
\[ 756 = \frac{V^2 - 6V}{20} \]

Now, multiply both sides by 20:
\[ 15120 = V^2 - 6V \]

Rearranging gives us:
\[ V^2 - 6V - 15120 = 0 \]

### Step 7: Factor or use the quadratic formula
We can use the quadratic formula \( V = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -6, c = -15120 \):
\[ V = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-15120)}}{2 \cdot 1} \]
\[ V = \frac{6 \pm \sqrt{36 + 60480}}{2} \]
\[ V = \frac{6 \pm \sqrt{60516}}{2} \]
Calculating the square root:
\[ \sqrt{60516} \approx 246 \]
Thus:
\[ V = \frac{6 \pm 246}{2} \]

Calculating the two possible values:
1. \( V = \frac{252}{2} = 126 \)
2. \( V = \frac{-240}{2} = -120 \) (not possible since speed cannot be negative)

### Conclusion
The usual speed of the car is \( V = 126 \) km/h.