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📝 How Solve Uniform Motion Applications (14 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 14 questions available

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What is How Solve Uniform Motion Applications?

Definition:
Uniform motion applications are word problems where objects move at constant speed along a straight line, and they are solved using the formula $D = rt$ (distance equals rate times time), with problems involving scenarios like catching up, traveling in opposite directions, or round trips, requiring setting up equations based on the relationships between distances, rates, and times.

Working:
To solve these problems, first identify what is given and what is asked, create a table with columns for distance, rate, and time, fill in the known values, use variables for unknowns, and set up equations based on the scenario (e.g., same distance for catching up, total distance for opposite directions, or equal times for round trips), then solve the resulting equation(s) using algebraic methods.

Example:
Two cars start from the same point and travel in opposite directions, one at 60 mph and the other at 40 mph; to find when they are 300 miles apart, set $60t + 40t = 300$ , so $100t = 300$ , and $t = 3$ hours, meaning after 3 hours they are 300 miles apart.

Reason:
Uniform motion applications are widely used in real life, including travel planning, logistics, and physics, and they help students apply the distance-rate-time relationship, which is a foundational concept in algebra and everyday problem-solving.

Easy

Medium

Hard

📝 All How Solve Uniform Motion Applications MCQs

Q1. Two cyclists start 18 km apart and ride toward each other. Their speeds are 6 km/h and 3 km/h. How long after starting will they meet?

A.1.5 hours ✅

B.2 hours

C.3 hours

D.6 hours

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Because the cyclists move toward each other, their closing speed is $6+3=9$ km/h. The meeting time is therefore $18/9=2$ hours, so option B is correct.

Q2. A car travels at a constant speed of 72 km/h. Which expression gives its distance $d$ , in kilometers, after $t$ hours?

d=72+t

d=72t

✅

d=t/72

d=72-t

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For uniform motion, distance equals constant speed multiplied by elapsed time. Since the speed is 72 km/h and time is $t$ hours, the appropriate model is $d=72t$ .

Q3. A delivery van leaves a warehouse at 8:00 a.m. traveling at 50 km/h. A second van leaves the same warehouse at 9:00 a.m. at 70 km/h on the same route. When will the second van catch the first?

A.10:30 a.m.

B.11:00 a.m. ✅

C.11:30 a.m.

D.12:00 noon

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The first van gains a one-hour head start and travels 50 km. The second van gains on it at $70-50=20$ km/h, requiring $50/20=2.5$ hours after 9:00 a.m., or 11:30 a.m. Therefore option C is correct.

Q4. Two trains leave stations 420 km apart at the same time and travel toward each other. One travels at 80 km/h and the other at 60 km/h. Which equation correctly models their meeting time $t$ in hours?

80t-60t=420

80t+60t=420

✅

420t=80+60

80+60+t=420

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Since both trains reduce the separation simultaneously, their distances add to 420 km. Thus $80t+60t=420$ , giving $140t=420$ and $t=3$ hours. The equation in option B correctly represents the situation.

Q5. A runner travels 12 km at a constant speed. If the runner increases the speed by 2 km/h, the trip takes 30 minutes less. What was the original speed?

A.4 km/h

B.5 km/h

C.6 km/h ✅

D.8 km/h

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Let the original speed be $v$ . Then $12/v-12/(v+2)=0.5$ . Multiplying through gives $24=0.5v(v+2)$ , so $v^2+2v-48=0$ . The positive solution is $v=6$ km/h.

Q6. A boat travels 24 km downstream in 2 hours and returns the same distance upstream in 3 hours. Assuming constant water and boat speeds, what is the boat's speed in still water?

A.8 km/h

B.10 km/h ✅

C.12 km/h

D.14 km/h

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The downstream speed is $24/2=12$ km/h and upstream speed is $24/3=8$ km/h. If $b$ is boat speed and $c$ is current speed, then $b+c=12$ and $b-c=8$ . Adding gives $2b=20$ , so $b=10$ km/h.

Q7. A student says, 'If two cars travel toward each other at 55 km/h and 65 km/h, their meeting time can be found by dividing the distance between them by 65 km/h.' What is the main error?

A.The faster car should be ignored completely

B.The speeds should be subtracted because the cars move toward each other

C.The closing speed should be the sum of their speeds ✅

D.Time should always be multiplied by speed

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: When two objects move toward one another, both contribute to reducing the separation. Therefore their relative or closing speed is $55+65=120$ km/h, not simply 65 km/h.

Q8. A train travels 180 km at a constant speed. A student calculates the travel time as $180\times90=16200$ hours when the speed is 90 km/h. Which correction is appropriate?

A.Divide distance by speed to obtain 2 hours ✅

B.Multiply speed by distance to obtain 2 hours

C.Subtract speed from distance to obtain 90 hours

D.Divide speed by distance to obtain 2 hours

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The student used multiplication when the relationship requires $t=d/v$ . Substituting $d=180$ km and $v=90$ km/h gives $t=180/90=2$ hours, with units confirming the calculation.

Q9. On a distance-versus-time graph, a vehicle is represented by a straight line rising steadily from $(0,0)$ to $(4,240)$ , where time is in hours and distance is in kilometers. What does the slope represent?

A.The total travel time

B.The vehicle's constant speed of 60 km/h ✅

C.The vehicle's acceleration of 240 km/h

D.The remaining distance after 4 hours

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The slope of a distance-time graph represents speed. Here the slope is $240/4=60$ km/h, indicating constant uniform motion at 60 km/h throughout the four-hour interval.

Q10. A courier must travel 150 km. For the first 2 hours, the courier travels at 60 km/h. What constant speed is required during the remaining distance to finish the trip in 3.5 hours total?

A.50 km/h

B.60 km/h

C.70 km/h ✅

D.80 km/h

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: During the first two hours, the courier covers $60(2)=120$ km, leaving 30 km. The remaining time is $3.5-2=1.5$ hours, so the required speed is $30/1.5=20$ km/h, making none of the listed options correct. Thus the scenario exposes an inconsistent answer set; the correct mathematical result is 20 km/h.

Q11. A graph shows two objects moving along the same straight route. Object A has position $20+40t$ , while Object B has position $140+10t$ , with position in kilometers and time in hours. When do their paths intersect?

A.2 hours

B.3 hours

C.4 hours ✅

D.5 hours

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: At the intersection, their positions are equal: $20+40t=140+10t$ . Rearranging gives $30t=120$ , so $t=4$ hours. Both objects are therefore at the same position after four hours.

Q12. A bus and a car start from the same point in opposite directions. The bus travels at 45 km/h and the car at 55 km/h. After 2.5 hours, how far apart are they?

A.25 km

B.100 km

C.137.5 km

D.250 km ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Because the vehicles travel in opposite directions, their separation increases at $45+55=100$ km/h. After 2.5 hours, the distance between them is $100(2.5)=250$ km.

Q13. A rescue team drives 96 km to an emergency location. Its planned speed is 48 km/h, but road conditions reduce the speed to 40 km/h. How much additional travel time results?

A.12 minutes

B.18 minutes

C.24 minutes ✅

D.30 minutes

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: At 48 km/h, the trip requires $96/48=2$ hours. At 40 km/h, it requires $96/40=2.4$ hours. The difference is $0.4$ hour, which equals 24 minutes.

Q14. Two hikers start 30 km apart and walk toward each other. One walks at 44 km/h, while the other walks at 66 km/h. A student claims they meet after 5 hours because $30/6=5$ . Which reasoning best identifies the flaw?

A.Only the slower hiker contributes to closing the distance

B.The hikers' speeds must be multiplied

C.Both hikers reduce the separation, so their speeds must be added ✅

D.The distance should be converted to meters first

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The student's calculation treats only one hiker as closing the gap. Since both move toward each other, the separation decreases at $4+6=10$ km/h, so they meet after $30/10=3$ hours.

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