If Greg drives at 40miles an hour and is 3 minutes late, and when he drives 60 miles he is 3 minutes early, and he always leaves home at 8 in the morning, how fast does he have to drive to get there just in time

We can solve this problem by establishing relationships between Greg's driving speeds, the time differences, and the distance he travels.

Let:
- \( d \) be the distance Greg has to travel (in miles).
- \( t \) be the time (in hours) it takes for Greg to arrive on time.

### Step 1: Express the time taken at different speeds

1. **When Greg drives at 40 miles per hour**:
   - Time taken = \( \frac{d}{40} \) hours.
   - He is 3 minutes late, which is equivalent to \( \frac{3}{60} = 0.05 \) hours.
   - So, the equation becomes:  
     \[
     \frac{d}{40} = t + 0.05
     \]

2. **When Greg drives at 60 miles per hour**:
   - Time taken = \( \frac{d}{60} \) hours.
   - He is 3 minutes early, which is \( \frac{3}{60} = 0.05 \) hours early.
   - The equation becomes:
     \[
     \frac{d}{60} = t - 0.05
     \]

### Step 2: Set up a system of equations

We now have the system of two equations:
1. \( \frac{d}{40} = t + 0.05 \)
2. \( \frac{d}{60} = t - 0.05 \)

### Step 3: Solve the system of equations

To eliminate \( t \), we subtract the second equation from the first:

\[
\frac{d}{40} - \frac{d}{60} = (t + 0.05) - (t - 0.05)
\]

Simplifying both sides:

\[
\frac{d}{40} - \frac{d}{60} = 0.1
\]

Find a common denominator for the left side:

\[
\frac{3d - 2d}{120} = 0.1
\]
\[
\frac{d}{120} = 0.1
\]
\[
d = 12
\]

So, the distance Greg travels is 12 miles.

### Step 4: Calculate the time to arrive just on time

We know the distance \( d = 12 \) miles. Now we can find the time \( t \).

Using the equation \( \frac{d}{40} = t + 0.05 \):

\[
\frac{12}{40} = t + 0.05
\]
\[
0.3 = t + 0.05
\]
\[
t = 0.25 \text{ hours} = 15 \text{ minutes}
\]

### Step 5: Determine the required speed

To arrive just on time, Greg needs to cover 12 miles in 15 minutes (or 0.25 hours). The speed is:

\[
\text{Speed} = \frac{d}{t} = \frac{12}{0.25} = 48 \text{ miles per hour}
\]

### Final Answer:
Greg must drive **48 miles per hour** to arrive exactly on time.

---

Would you like more details, or do you have any other questions?

Here are five related questions to consider:
1. How would the answer change if Greg’s destination were 24 miles away instead of 12?
2. What if Greg leaves home 5 minutes later than usual, how does it affect his driving speed?
3. How much earlier or later would Greg be if he drives at 50 miles per hour instead of 48?
4. If Greg had to make a stop that took 5 minutes, how fast should he drive to still arrive on time?
5. What is the relationship between speed and time in general?

**Tip:** Always check units (miles, minutes, hours) to avoid calculation errors when solving word problems.

If Greg drives at 40 miles an hour and is 3 minutes late, and when he drives 60 miles an hour he is 3 minutes early, and he always leaves home at 8 in the morning, how fast does he have to drive to get there just in time?

Learn how to calculate the speed Greg must drive to arrive just on time. The problem uses key mathematical concepts of linear equations and time-distance relationships, providing a step-by-step solution suitable for students in Grades 9-11.

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