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

Answer

To solve the problem, we need to find the usual speed of the car (let's denote it as \( s \) km/h). Here’s how we can approach the solution step by step:

### Step 1: Understand the problem
The car leaves 3 minutes early and has to slow down by 6 km/h to reach a destination that is 126 km away. We need to find the usual speed of the car.

### Step 2: Convert time to hours
Since speed is in km/h, we need to convert the 3 minutes into hours:
\[ 3 \text{ minutes} = \frac{3}{60} \text{ hours} = \frac{1}{20} \text{ hours} \]

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

### Step 4: Calculate the time taken at usual speed
The time taken to cover 126 km at the usual speed \( s \) is:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{126}{s} \text{ hours} \]

### Step 5: Calculate the time taken at reduced speed
The time taken to cover 126 km at the reduced speed \( s - 6 \) is:
\[ \text{Time} = \frac{126}{s - 6} \text{ hours} \]

### Step 6: Relate the times
Since the car leaves 3 minutes (or \( \frac{1}{20} \) hours) early, the time taken at the reduced speed must be equal to the time taken at the usual speed minus \( \frac{1}{20} \) hours:
\[ \frac{126}{s - 6} = \frac{126}{s} - \frac{1}{20} \]

### Step 7: Solve the equation
To solve for \( s \), we can first eliminate the fractions by multiplying through by \( s(s - 6) \) (the common denominator):
\[ 126s = 126(s - 6) - \frac{s(s - 6)}{20} \]

Expanding and simplifying:
\[ 126s = 126s - 756 - \frac{s^2 - 6s}{20} \]

Now, multiply through by 20 to eliminate the fraction:
\[ 2520s = 2520s - 15120 - s^2 + 6s \]

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

### Step 8: Use the quadratic formula
Using the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -6, c = -15120 \):
\[ s = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-15120)}}{2 \cdot 1} \]
\[ s = \frac{6 \pm \sqrt{36 + 60480}}{2} \]
\[ s = \frac{6 \pm \sqrt{60516}}{2} \]
Calculating \( \sqrt{60516} \approx 246 \):
\[ s = \frac{6 \pm 246}{2} \]

### Step 9: Calculate the values
Calculating the two possible values:
1. \( s = \frac{252}{2} = 126 \)
2. \( s = \frac{-240}{2} = -120 \) (not valid since speed cannot be negative)

Thus, the usual speed of the car is:
\[ \boxed{126 \text{ km/h}} \]