Jack’s father drives to work in 30 minutes when driving at his usual speed

Jack’s father drives to work in 30 minutes when driving at his usual speed. When traffic is bad, he drives 30 miles per hour slower and the trip takes one hour longer. How far is Jack’s father’s work from home?

The correct answer and explanation is:

To find how far Jack’s father’s work is from home, let’s use the information given and set up an equation.

Let:

• dd = distance from home to work (in miles)
• rr = usual speed (in miles per hour)

From the problem:

1. Usual trip:
   • Time: 30 minutes = 0.5 hours
   • Speed: rr
   • So, d=r×0.5d = r \times 0.5
2. Bad traffic trip:
   • Speed: r−30r – 30
   • Time: 1 hour longer than usual → 0.5 + 1 = 1.5 hours
   • So, d=(r−30)×1.5d = (r – 30) \times 1.5

Set up the equation:

Since both expressions equal the distance dd, we equate them: r×0.5=(r−30)×1.5r \times 0.5 = (r – 30) \times 1.5

Multiply both sides to eliminate decimals: 0.5r=1.5(r−30)0.5r = 1.5(r – 30) 0.5r=1.5r−450.5r = 1.5r – 45

Subtract 0.5r0.5r from both sides: 0=1.0r−450 = 1.0r – 45 1.0r=451.0r = 45 r=45r = 45

Find the distance:

Now plug r=45r = 45 into the distance formula: d=0.5×45=22.5 milesd = 0.5 \times 45 = 22.5 \text{ miles}

✅ Final Answer: 22.5 miles

Explanation (300 words):

This problem involves understanding the relationship between speed, distance, and time, using the formula: Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}

Jack’s father normally drives to work in 30 minutes, which is 0.5 hours, at his usual speed, which we call rr. So the normal distance to work is: d=r×0.5d = r \times 0.5

When traffic is bad, his speed decreases by 30 miles per hour, making it r−30r – 30, and the trip takes an extra hour, totaling 1.5 hours. The distance remains the same, so we write: d=(r−30)×1.5d = (r – 30) \times 1.5

Since both expressions equal the same distance dd, we set them equal and solve for rr. This gives r=45r = 45 miles per hour. That means under normal conditions, he drives 45 mph.

To find the distance, we substitute back into the formula: d=0.5×45=22.5 milesd = 0.5 \times 45 = 22.5 \text{ miles}

This answer makes sense when we check it:

• Normal trip: 22.5÷45=0.522.5 \div 45 = 0.5 hours
• Bad traffic: 22.5÷15=1.522.5 \div 15 = 1.5 hours (since 45 – 30 = 15 mph)

The math checks out perfectly. Therefore, Jack’s father’s work is 22.5 miles from home.