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📝 Catching up word formula algebra (12 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 12 questions available

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What is Catching up word formula algebra?

Definition:
Catching up word problems involve two moving objects, where one starts later or moves faster and eventually catches up to the other, and the solution uses the formula $D = rt$ with the key relationship that the distances traveled by both are equal at the catching-up point, often requiring setting up an equation with the difference in start times or rates.

Working:
To solve catching up problems, create a table with rows for each object, columns for distance, rate, and time, express the time for the second object as the first object's time minus the time delay (or plus if it starts later), then set the distances equal, solve for the variable, and check the reasonableness of the answer; the general equation is $r_1 t = r_2 (t - \text{delay})$ if they start at different times.

Example:
A car leaves at 50 mph, and 2 hours later, another car leaves from the same point at 70 mph; to find when the second car catches up, set $50t = 70(t - 2)$ , solving gives $50t = 70t - 140$ , so $-20t = -140$ , $t = 7$ hours, meaning the first car has traveled 7 hours, and the second car travels 5 hours to catch up.

Reason:
Catching up problems are common in real-life scenarios like traffic, racing, and logistics, and they help develop algebraic reasoning and understanding of motion, making them a key application in algebra and physics.

Easy

Medium

Hard

📝 All Catching up word formula algebra MCQs

Q1. A cyclist travels at 1818 km/h while another cyclist ahead travels in the same direction at 1212 km/h. Which relative-speed expression correctly describes how quickly the first cyclist reduces the gap?

18+12=30

km/h

18-12=6

km/h ✅

12-18=6

km/h

18\times12=216

km/h

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When two objects move in the same direction, the faster object closes the gap at the difference of their speeds. Thus the relative speed is $18-12=6$ km/h, not the sum of their speeds.

Q2. A runner moving at 99 m/s is chasing another runner moving at 77 m/s. If the initial separation is 3030 m, which equation represents the catch-up time $t$ ?

9t+7t=30

9t-7t=30

✅

7t-9t=30

16t=30

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Because both runners move in the same direction, their separation decreases by $9-7=2$ meters each second. Therefore, $2t=30$ , or equivalently $9t-7t=30$ , correctly models the catch-up time.

Q3. A delivery van travels at 6060 km/h and follows a truck traveling at 4545 km/h. Why is 1515 km/h the useful speed for determining how fast the van closes the gap?

A.It is the truck's speed after acceleration

B.It is the average of the two speeds

C.It is the difference between the speeds in the same direction ✅

D.It is the distance traveled by the van each hour

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For motion in the same direction, both vehicles advance along the road simultaneously. The gap changes according to their speed difference, $60-45=15$ km/h, which is the closing or relative speed.

Q4. A car is 2424 km ahead of a motorcycle. The car travels at 5050 km/h, while the motorcycle travels at 6262 km/h. A student says the motorcycle catches the car in $24/62$ hours. What is the key flaw?

A.The student should multiply the speeds

B.The student ignores the car's continued motion ✅

C.The student should use the sum of the speeds

D.The student should use the car's speed only

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The car does not remain stationary while being chased. It continues moving at 50 km/h, so the motorcycle closes the gap at $62-50=12$ km/h. The catch-up time is therefore $24/12=2$ hours.

Q5. A bus leaves a station at 4040 km/h. Thirty minutes later, a car leaves the same station in the same direction at 7070 km/h. How long after the car starts will it catch the bus?

A.0.50 h

B.0.67 h

C.1.00 h ✅

D.1.33 h

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: During the first 0.5 hour, the bus gains a $40(0.5)=20$ km lead. The car closes this gap at $70-40=30$ km/h, requiring $20/30=2/3$ hour, so the answer is approximately 0.67 h, not 1.00 h.

Q6. A police vehicle travels at 8080 km/h and begins chasing a vehicle 1515 km ahead traveling at 6565 km/h. After 2020 minutes, has the police vehicle caught the other vehicle?

A.Yes, because it travels 26.67 km

B.Yes, because the relative distance is 15 km

C.No, because it closes only 5 km ✅

D.No, because both vehicles travel the same distance

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The relative speed is $80-65=15$ km/h. In 20 minutes, or $1/3$ hour, the police vehicle closes $15(1/3)=5$ km. Since the original gap is 15 km, 10 km remains.

Q7. A train traveling at 9090 km/h is 4545 km behind another train traveling at 7575 km/h. If both continue at constant speeds, how long will the faster train need to catch the slower train?

A.1.5 hours

B.2 hours

C.3 hours ✅

D.6 hours

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The gap closes at $90-75=15$ km/h. With a 45-km initial separation, the required time is $45/15=3$ hours. The larger distance traveled by both trains does not change the relative-motion calculation.

Q8. A student calculates a catch-up time using $t=\frac{d}{v_1+v_2}$ for two cars moving in the same direction. Which situation would make this calculation appropriate instead?

A.The cars have identical speeds

B.The cars move in opposite directions toward each other ✅

C.The faster car starts behind the slower car

D.The cars start at different times but move in the same direction

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Adding speeds gives the rate at which separation changes when two objects move toward one another in opposite directions. For same-direction chasing, the correct rate is the difference between their speeds.

Q9. A distance-versus-time graph shows two straight lines. Line A starts at 00 km and has slope 6060 km/h. Line B starts at 1515 km and has slope 4545 km/h. What does the intersection of the lines represent?

A.The average speed of the two objects

B.The starting position of Line B

C.The moment and position when the faster object catches the slower object ✅

D.The time when both objects stop

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: On a distance-versus-time graph, slope represents speed and an intersection means both objects have the same position at the same time. Since A has the greater slope but begins behind, the intersection represents catching up.

Q10. A cyclist travels 1010 km at 2020 km/h before a second cyclist starts from the same point at 3030 km/h. How long after the second cyclist starts will the second cyclist catch the first?

A.0.5 h

B.1 h ✅

C.1.5 h

D.2 h

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The first cyclist has a 10-km head start. The second cyclist gains on the first at $30-20=10$ km/h. Therefore, the catch-up time is $10/10=1$ hour after the second cyclist begins.

Q11. A runner is 100100 m ahead and runs at 55 m/s. A faster runner starts behind at 88 m/s. An incorrect solution says the catch-up time is $100/(8+5)$ . What result should replace it?

A.7.69 s

B.20 s

C.33.33 s ✅

D.40 s

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The runners move in the same direction, so their separation decreases at $8-5=3$ m/s. Starting 100 m apart, the catch-up time is $100/3\approx33.33$ seconds. The incorrect solution uses a sum appropriate to closing from opposite directions.

Q12. A bus has a 1212-km lead and travels at 4848 km/h. A car behind it travels at 7272 km/h. The car's driver claims that increasing the car's speed by 1212 km/h will cut the remaining catch-up time exactly in half. Is the claim correct?

A.Yes, because the car becomes twice as fast

B.Yes, because the relative speed doubles

C.No, because the new relative speed is 36 km/h rather than twice 24 km/h ✅

D.No, because the bus also doubles its speed

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Initially, the closing speed is $72-48=24$ km/h, giving a catch-up time of $12/24=0.5$ hour. After increasing the car to 84 km/h, the closing speed is 36 km/h, giving $12/36=1/3$ hour, not half of the original time.

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