Distance, Rate, and Time Word Problems Worksheet

Let's solve each of these Distance, Rate, and Time word problems step by step. The formula we'll use is:

$$
\text{Distance} = \text{Rate} \times \text{Time} \quad \text{(or } D = R \times T\text{)}
$$

We’ll go through each problem carefully.

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1) Sara left downtown Portland, and three hours later, Sam left going 43 mph faster to catch up. After another two hours, Sam caught up. Find Sara's average speed.

- Let $ x $ be Sara’s speed (in mph).
- Sam’s speed = $ x + 43 $
- Sam started 3 hours after Sara.
- Sam caught up after 2 more hours, so Sam traveled for 2 hours.
- In that time, Sara had been traveling for $ 3 + 2 = 5 $ hours.

Since they both traveled the same distance when Sam caught up:
$$
\text{Sara's distance} = \text{Sam's distance}
$$
$$
x \cdot 5 = (x + 43) \cdot 2
$$
$$
5x = 2x + 86
$$
$$
3x = 86
$$
$$
x = \frac{86}{3} \approx 28.67 \text{ mph}
$$

✔ Answer: Sara's average speed is $ \boxed{28.\overline{6}} $ mph or $ \boxed{\frac{86}{3}} $ mph.

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2) Nancy traveled to Durham by car. Going there took three hours, and the return trip lasted eight hours. Nancy averaged a speed of forty-two mph while returning. Find the average speed of the trip there.

- Return trip: time = 8 hours, speed = 42 mph
- So, distance one way = $ 42 \times 8 = 336 $ miles
- Going there: time = 3 hours, distance = 336 miles
- Speed there = $ \frac{336}{3} = 112 $ mph

✔ Answer: Average speed going there = $ \boxed{112} $ mph

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3) A train left for Burbank, and 3 hours later, a car traveling 68 mph tried catching up to the train. After 9 hours, the car caught up. What was the train's average speed?

- Car traveled 9 hours at 68 mph → distance = $ 68 \times 9 = 612 $ miles
- Train had a 3-hour head start, so it traveled $ 3 + 9 = 12 $ hours
- Distance = 612 miles (same as car)
- Train’s speed = $ \frac{612}{12} = 51 $ mph

✔ Answer: Train’s average speed = $ \boxed{51} $ mph

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4) Mary left Atlanta with a speed of forty-five mph. Fred also left at the same time in the opposite direction at a speed of sixty mph. Find how many hours Fred must travel before they are three hundred and eighteen miles apart.

- They’re moving in opposite directions → their speeds add
- Combined speed = $ 45 + 60 = 105 $ mph
- Distance apart = 318 miles
- Time = $ \frac{318}{105} = 3.02857... $

Let’s simplify:
$$
\frac{318}{105} = \frac{106}{35} \approx 3.0286 \text{ hours}
$$

But let's check if it reduces:
- $ 318 \div 3 = 106 $, $ 105 \div 3 = 35 $ → $ \frac{106}{35} $

✔ Answer: $ \boxed{\frac{106}{35}} $ hours or approximately $ \boxed{3.03} $ hours

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5) Keith left the city traveling at 83 mph, while, at the same time, Sara left the city going the opposite direction at a speed of 64 mph. Find the time Keith traveled before the two were 153 miles apart.

- Opposite directions → combined speed = $ 83 + 64 = 147 $ mph
- Distance = 153 miles
- Time = $ \frac{153}{147} = \frac{51}{49} \approx 1.04 $ hours

✔ Answer: $ \boxed{\frac{51}{49}} $ hours or about $ \boxed{1.04} $ hours

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6) Sally left the city for vacation. Sandy left six hours later going eighty-two mph faster to catch up. After five hours Sandy caught up. What was Sally's average speed?

- Let $ x $ = Sally’s speed
- Sandy’s speed = $ x + 82 $
- Sandy traveled 5 hours → distance = $ 5(x + 82) $
- Sally traveled $ 6 + 5 = 11 $ hours → distance = $ 11x $
- Equal distances:
$$
11x = 5(x + 82)
$$
$$
11x = 5x + 410
$$
$$
6x = 410
$$
$$
x = \frac{410}{6} = \frac{205}{3} \approx 68.33 \text{ mph}
$$

✔ Answer: Sally’s average speed = $ \boxed{\frac{205}{3}} $ mph or $ \boxed{68.\overline{3}} $ mph

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7) Alyssa left Durham traveling 50 mph. Jessica, to catch up, left some time later driving at 59 mph. Jessica caught up after 4 hours. How long was Alyssa driving before Jessica caught up?

- Jessica drove 4 hours at 59 mph → distance = $ 59 \times 4 = 236 $ miles
- Alyssa was already driving when Jessica started, so she drove longer
- Let $ t $ = time Alyssa drove before Jessica caught up
- Then: $ 50t = 236 $
$$
t = \frac{236}{50} = 4.72 \text{ hours}
$$

But wait — this would mean Alyssa drove 4.72 hours, but Jessica only drove 4 hours. That makes sense.

Wait: Is this correct?

Yes: When Jessica catches up, both have traveled same distance.

So:
- Jessica: $ 59 \times 4 = 236 $ miles
- Alyssa: $ 50 \times t = 236 $
- $ t = \frac{236}{50} = 4.72 $ hours

So Alyssa had been driving for $ \boxed{4.72} $ hours before Jessica caught up.

But let’s express as fraction:
$$
\frac{236}{50} = \frac{118}{25} = 4 \frac{18}{25} \text{ hours}
$$

✔ Answer: $ \boxed{4.72} $ hours or $ \boxed{\frac{118}{25}} $ hours

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8) A cargo plane flew from the US across the Atlantic at two hundred and ninety-one mph, and flew back to the US at three hundred and eleven mph. Given that the first trip took two hours longer, how long was the return trip?

Let:
- $ t $ = time of return trip (in hours)
- Then, outbound trip time = $ t + 2 $
- Distance is same both ways

So:
$$
\text{Outbound distance} = \text{Return distance}
$$
$$
291(t + 2) = 311t
$$
$$
291t + 582 = 311t
$$
$$
582 = 20t
$$
$$
t = \frac{582}{20} = 29.1 \text{ hours}
$$

✔ Answer: Return trip was $ \boxed{29.1} $ hours

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9) A truck and van left from Portland in opposite directions. The truck traveled for 6 hours at 48 mph. The vehicles were 586 miles apart. Find the van's average speed.

- Truck distance = $ 48 \times 6 = 288 $ miles
- Total distance apart = 586 miles
- So van traveled $ 586 - 288 = 298 $ miles
- Van traveled for 6 hours (same time)
- Van speed = $ \frac{298}{6} = \frac{149}{3} \approx 49.67 $ mph

✔ Answer: Van’s average speed = $ \boxed{\frac{149}{3}} $ mph or $ \boxed{49.\overline{6}} $ mph

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10) A plane set off to Burbank at a speed of two hundred and seven mph. On the return flight of ten hours, the plane cruised at two hundred and sixty-seven mph. How many hours long was the flight to Burbank?

- Return flight: 10 hours at 267 mph → distance = $ 267 \times 10 = 2670 $ miles
- Outbound flight: same distance, speed = 207 mph
- Time = $ \frac{2670}{207} $

Simplify:
$$
\frac{2670}{207} = \frac{2670 \div 3}{207 \div 3} = \frac{890}{69} \approx 12.8986 \text{ hours}
$$

Check:
- $ 207 \times 12.8986 \approx 2670 $? Yes.

Let’s divide:
$$
2670 \div 207 = 12.8986...
$$

Or as fraction: $ \frac{890}{69} $ hours

✔ Answer: Flight to Burbank was $ \boxed{\frac{890}{69}} $ hours or approximately $ \boxed{12.9} $ hours

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ \frac{86}{3} $ mph ≈ 28.67 mph |
| 2 | 112 mph |
| 3 | 51 mph |
| 4 | $ \frac{106}{35} $ hours ≈ 3.03 hrs |
| 5 | $ \frac{51}{49} $ hours ≈ 1.04 hrs |
| 6 | $ \frac{205}{3} $ mph ≈ 68.33 mph |
| 7 | $ \frac{118}{25} $ hours = 4.72 hrs |
| 8 | 29.1 hours |
| 9 | $ \frac{149}{3} $ mph ≈ 49.67 mph |
| 10 | $ \frac{890}{69} $ hours ≈ 12.9 hrs |

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