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Motion Problems

Motion problems involve objects moving at constant speed and use relationships between distance, rate (speed), and time.

Key Formula:

$$ \text{Distance} = \text{Rate} \times \text{Time} $$

Variable Notation:

Common Motion Scenarios:

Use the distance formula to write equations for each object or direction, then solve based on the relationship given in the problem.

Variable Speed

In problems involving rivers or flowing water, the motion of a boat or swimmer is affected by the speed of the current. The total effective speed depends on whether the motion is going with or against the current.

Key Relationships:

Let the variables be:

Effective Speeds:

Once you know the effective rate, use the distance formula:

$$ \text{Distance} = \text{Rate} \times \text{Time} $$

This type of problem often involves comparing times or distances for round trips or upstream/downstream travel.

Concept Concept Concept Concept Concept Concept Concept Concept Concept

Problem:

Two pals A and B run at constant speeds along a circular track 1350m in circumference. Running in opposite directions they meet every 3 minutes, while running in the same direction they are together every 27 minutes. Determine their speeds in km/hr. Ans. A=15kph B=12kph

Motion Problems | Algebra – Problem 1: – Diagram Motion Problems | Algebra – Problem 1: – Diagram Motion Problems | Algebra – Problem 1: – Diagram

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Motion Problems | Algebra – Problem 1: – Diagram Motion Problems | Algebra – Problem 1: – Diagram Motion Problems | Algebra – Problem 1: – Diagram Motion Problems | Algebra – Problem 1: – Diagram

Problem:

An airplane travels 500km against the wind in one hour and 45 minutes. Travelling the same distance with the wind, the airplane consumed one hour and 15 minutes. Find the velocity of the wind in kph. Ans. 57.14kph

Motion Problems | Algebra – Problem 2: – Diagram Motion Problems | Algebra – Problem 2: – Diagram Motion Problems | Algebra – Problem 2: – Diagram

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Motion Problems | Algebra – Problem 2: – Diagram Motion Problems | Algebra – Problem 2: – Diagram Motion Problems | Algebra – Problem 2: – Diagram Motion Problems | Algebra – Problem 2: – Diagram

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Motion Problems | Algebra – Problem 3: – Diagram Motion Problems | Algebra – Problem 3: – Diagram Motion Problems | Algebra – Problem 3: – Diagram Motion Problems | Algebra – Problem 3: – Diagram

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q61

MSTE - Algebra / Motion Problems / Engr. Janclyde Espinosa (Clidez)

Two pals A and B run at constant speeds along a circumference track 1350m in circumference. Running in opposite directions they meet every 3 minutes, while running in the same direction they are together every 27 minutes. Determine their speeds in km/hr.

What is the speed of A?

  1. 15
  2. 12
  3. 17
  4. 13

What is the speed of B?

  1. 12
  2. 15
  3. 13
  4. 17

Part 1.

Let speeds be $a$ and $b$ in m/min. Opposite directions:
$a+b=1350/3=450$
Same direction:
$a-b=1350/27=50$
Solving gives $a=250$ m/min. Convert to km/hr:
$250(60)/1000=15$
$\boxed{15\text{ km/hr}}$

Part 2.

From $a+b=450$ and $a-b=50$, $b=200$ m/min. Convert:
$200(60)/1000=12$
$\boxed{12\text{ km/hr}}$

Question Bank: q261

MSTE - Algebra / Motion Problems / Engr. Janclyde Espinosa (Clidez)

The speed of a Ranger boat in still water is 32 mph. If the boat travels 72 miles upstream at the same time that it takes to travel 120 miles downstream, find the current of the stream.

Answer:

  1. 8 mph
  2. 20 mph
  3. 16 mph
  4. 24 mph
Let the current speed be $c$. Upstream speed is $32-c$ and downstream speed is $32+c$. Equal times give:
$\frac{72}{32-c}=\frac{120}{32+c}$
$72(32+c)=120(32-c)$
$192c=1536$
$\boxed{c=8\text{ mph}}$

Question Bank: q670

MSTE - Algebra / Motion Problems / Engr. Janclyde Espinosa (Clidez)

Cars A and B leave a town at the same time and travel in opposite directions. Car A travels at 40 kph and car B at 60 kph. How long will the two cars be 350 km apart?

  1. 3.5 hrs
  2. 5.5 hrs
  3. 2.5 hrs
  4. 4 hrs
The cars separate at their combined speed:
$40+60=100$ kph
Time for 350 km separation:
$t=350/100$
$\boxed{3.5\text{ hrs}}$

Question Bank: q700

MSTE - Algebra / Motion Problems / Engr. Janclyde Espinosa (Clidez)

An army of troops is marching along a road at 5 kph. A messenger on horseback was sent from the front to the rear of the column and returns immediately back. The total time taken is 10 minutes. Assuming the messenger rides at the rate of 10 kph, determine the length of the column.

  1. 625 m
  2. 526 m
  3. 1 km
  4. 854 m
Let the column length be $L$ km and messenger speed be 10 kph while troops move at 5 kph. Time front-to-rear:
$t_1=\frac{L}{10+5}$
Time rear-to-front:
$t_2=\frac{L}{10-5}$
Total time is 10 min = $1/6$ hr:
$\frac{L}{15}+\frac{L}{5}=\frac{1}{6}$
$\frac{4L}{15}=\frac{1}{6}$
$L=0.625$ km
$\boxed{625\text{ m}}$

Question Bank: q704

MSTE - Algebra / Motion Problems / Engr. Janclyde Espinosa (Clidez)

A man drives from station A to station B 60 km away at an average speed of 30 kph and returns to A at an average speed of 20 kph. What was the average speed of the man in the whole journey?

  1. 24 kph
  2. 20 kph
  3. 25 kph
  4. 22 kph
Average speed for the whole round trip is total distance over total time:
$d=60+60=120$ km
$t=60/30+60/20=2+3=5$ hr
$v_{avg}=120/5$
$\boxed{24\text{ kph}}$

Question Bank: q708

MSTE - Algebra / Motion Problems / Engr. Janclyde Espinosa (Clidez)

Two men are walking toward each other along a railway. A freight train overtakes one of them in 20 seconds and exactly 10 minutes later meets the other man coming from the opposite direction. The train passes this second man in 18 seconds. How long after the train has passed the second man will the two men meet?

  1. 1 hr 32 min 42 sec
  2. 1 hr 42 min 32 sec
  3. 1 hr 24 min 23 sec
  4. 1 hr 23 min 24 sec

Solution pending in psadquestions/q708.json.

Question Bank: q726

MSTE - Algebra / Motion Problems / Engr. Janclyde Espinosa (Clidez)

A battalion 20 miles long advances 20 miles. During this time, a messenger on a horse travels from the rear of the battalion to the front, immediately turns around, and ends up precisely at the rear of the battalion upon completion of the 20-mile journey. How far has the messenger traveled?

  1. 48.28 miles
  2. 40 miles
  3. 50 miles
  4. 52.42 miles
Let battalion speed be $v$ and horse speed be $u$. The battalion advances 20 mi while its length is 20 mi, so time is $20/v$. Messenger travel time rear-to-front is $20/(u-v)$ and front-to-rear is $20/(u+v)$. Thus:
$\frac{20}{u-v}+\frac{20}{u+v}=\frac{20}{v}$
This gives $u/v=1+\sqrt2$. Messenger distance is $u(20/v)=20(1+\sqrt2)$.
$\boxed{48.28\text{ miles}}$

Question Bank: t14

MSTE - Algebra / Motion Problems / Civil Engineering Refresher

David's route is 80 miles and Keith's is 100 miles. Keith averages 10 mph more than David and finishes 10 minutes before David. What is David's speed?

  1. 30 mph
  2. 40 mph
  3. 25 mph
  4. 35 mph
Let $v$ = David's speed; Keith's $= v + 10$. Keith saves $10$ min $= \frac{1}{6}$ hr:
$\frac{80}{v} - \frac{100}{v+10} = \frac{1}{6}$
Testing $v = 30$: $\frac{80}{30} - \frac{100}{40} = 2.667 - 2.5 = \frac{1}{6}$ ✓
$\boxed{v = 30 \text{ mph}}$

Question Bank: t15

MSTE - Algebra / Motion Problems / Civil Engineering Refresher

A plane flies 465 miles with the wind and 345 miles against the wind in the same amount of time. If the wind speed is 20 mph, find the speed of the plane in still air.

  1. 135 mph
  2. 150 mph
  3. 125 mph
  4. 140 mph
Equal times: $\frac{465}{p+20} = \frac{345}{p-20}$
$465(p-20) = 345(p+20) \Rightarrow 120p = 16200$
$\boxed{p = 135 \text{ mph}}$

Question Bank: t177

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

A car travelled for $p$ hours with a velocity of magnitude $q$ kph, then travels the same direction for $r$ hours with velocity of magnitude $s$ kph. What is the average velocity during the travel?

  1. $(pq + rs)/(p + r)$
  2. $(pr + qs)/(p + r)$
  3. $(pr + qs)/(q + s)$
  4. $(pq + rs)/(q + s)$
Average velocity is total displacement divided by total time.
First displacement $=pq$; second displacement $=rs$.
Total time $=p+r$.
$v_{avg}=\frac{pq+rs}{p+r}$
$\boxed{\frac{pq+rs}{p+r}}$

Question Bank: t178

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

At 2:25 PM, an engineer and his girlfriend walks along Osmeña Boulevard for 30 minutes at a speed of 3 kph. They waited for a taxi for 10 minutes which brought them back to where they start at 3:15 PM. What was the average speed of the taxi?

  1. 4.75 kph
  2. 6.25 kph
  3. 9.00 kph
  4. 2.25 kph
They walked for 30 minutes at 3 kph, so distance from the start is $3(0.5)=1.5$ km.
After waiting 10 minutes, the taxi trip is from 3:05 PM to 3:15 PM, or $\frac{1}{6}$ hr.
Taxi speed $=\frac{1.5}{1/6}=9.00$ kph.
$\boxed{9.00\text{ kph}}$

Question Bank: t179

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

A boat takes 2/3 as much time to travel downstream from C to D, as to return. If the rate of the river's current is 8 kph, what is the speed of the boat in still water?

  1. 40 kph
  2. 38 kph
  3. 41 kph
  4. 39 kph
Let $v$ be the boat speed in still water. Downstream speed is $v+8$ and upstream speed is $v-8$.
For the same distance, downstream time is $\frac{2}{3}$ of upstream time:
$\frac{d}{v+8}=\frac{2}{3}\left(\frac{d}{v-8}\right)$
$3(v-8)=2(v+8)$
$v=40$ kph.
$\boxed{40\text{ kph}}$

Question Bank: t184

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

It takes 12 seconds for two trains to pass each other when moving in opposite directions. The faster train which is 130 m long is moving at 13.5 m/s. What is the speed of the slower train which is 120 m long?

  1. 7.33 m/s
  2. 7.67 m/s
  3. 8.33 m/s
  4. 8.67 m/s
When trains pass in opposite directions, relative speed equals total length divided by time.
Total length $=130+120=250$ m.
Relative speed $=\frac{250}{12}=20.833$ m/s.
Slower train speed $=20.833-13.5=7.33$ m/s.
$\boxed{7.33\text{ m/s}}$

Question Bank: t188

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

A train takes four hours and thirty minutes to travel from station A to station B. How long will the journey be if:

The train's speed is increased to 1.5 times its original speed.

  1. 2 hrs
  2. 3 hrs
  3. 4 hrs
  4. 5 hrs

The train's speed is increased to 3.75 times its original speed.

  1. 1.8 hrs
  2. 1.5 hrs
  3. 2.4 hrs
  4. 1.2 hrs

The train's speed is decreased to 0.75 times its original speed.

  1. 6 hrs
  2. 8 hrs
  3. 5 hrs
  4. 7 hrs

Part 1.

Original time is 4 hr 30 min $=4.5$ hr. For the same distance, time is inversely proportional to speed.
If speed becomes $1.5$ times original, new time $=\frac{4.5}{1.5}=3$ hr.
$\boxed{3\text{ hrs}}$

Part 2.

For the same distance, time varies inversely with speed.
New time $=\frac{4.5}{3.75}=1.2$ hr.
$\boxed{1.2\text{ hrs}}$

Part 3.

For the same distance, time varies inversely with speed.
If speed becomes $0.75$ of the original, new time $=\frac{4.5}{0.75}=6$ hr.
$\boxed{6\text{ hrs}}$

Question Bank: w15

MSTE - Algebra / Motion Problems / MSTE May 2019

Two trains going in opposite directions leave at the same time. One train travels 15 mph faster than the other. In 6 hours, the trains are 630 miles apart. Find the speed of the faster train.

  1. 60 mph
  2. 45 mph
  3. 75 mph
  4. 30 mph
Let the slower speed be $v$.
$6v + 6(v + 15) = 630$
$12v + 90 = 630 \Rightarrow v = 45\text{ mph}$
Faster train $= 45 + 15 = \boxed{60\text{ mph}}$
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