Motion Problems — Speed, Distance and Time

Motion problems are among the most common word problems, and they all rely on one simple relationship between speed, distance and time. Once you understand that relationship you can solve meeting problems, overtaking problems, and average-speed problems. On this page we will learn the fundamental formula, practise unit conversions, and develop a strategy for problems involving two moving objects.

Background and Basic Definitions

The central relationship in all motion problems is:

\[ v = \frac{d}{t} \]

where \(v\) is speed, \(d\) is distance, and \(t\) is time. Two useful rearrangements follow:

Distance: \( d = v \cdot t \)
Time: \( t = \frac{d}{v} \)

Unit conversion is a common source of errors: minutes must be converted to hours before substituting. For example, \(30\) minutes is \(0.5\) hours, \(45\) minutes is \(0.75\) hours, and one hour and a quarter is \(1.25\) hours.

Meeting problems (head-on): when two objects travel toward each other, the distance between them decreases at the sum of their speeds (closing speed). Meeting time = initial distance ÷ sum of speeds.

Overtaking problems (same direction): the gap closes at the difference of their speeds.

Average speed is not the arithmetic mean of the speeds but rather:

\[ \bar{v} = \frac{\text{total distance}}{\text{total time}} \]

Solution Steps

Step 1 — List the given values and convert all times to hours (or all speeds to the same units).
Step 2 — Identify what is being asked (\(v\), \(d\) or \(t\)) and choose the appropriate form of the formula.
Step 3 — For two-object problems, determine whether they travel toward each other (sum of speeds) or in the same direction (difference of speeds).
Step 4 — If the objects depart at different times, calculate how far the head-start object has travelled before the second one departs.
Step 5 — Substitute into the equation, solve, and for average speed always use total distance ÷ total time.
Step 6 — Check for reasonableness: the answer is positive, the units are correct, and the magnitude makes sense.

Worked Examples

Example 1: Basic Speed Calculation

Problem: A bus travelled \(150\) km in two and a half hours. What was its average speed in km/h?

Solution:

Convert the time to hours: two and a half hours is \( 2.5 \) hours.
Use the formula \( v = \frac{d}{t} = \frac{150}{2.5} \).
Compute: \( \frac{150}{2.5} = 60 \).

Answer: The average speed was \( 60 \) km/h.

Example 2: Finding Time with Unit Conversion

Problem: A cyclist rode \(18\) km at a speed of \(24\) km/h. How long did the journey take (in minutes)?

Solution:

Use \( t = \frac{d}{v} = \frac{18}{24} \).
Compute: \( \frac{18}{24} = 0.75 \) hours.
Convert to minutes: \( 0.75 \times 60 = 45 \) minutes.

Answer: The journey took \( 45 \) minutes.

Example 3: Meeting Problem

Problem: Two cities are \(120\) km apart. A car leaves city \(A\) at \(60\) km/h, and simultaneously a car leaves city \(B\) toward it at \(40\) km/h. How long until they meet, and how far from \(A\)?

Solution:

The cars travel toward each other, so the closing speed is the sum: \( 60 + 40 = 100 \) km/h.
Meeting time: \( t = \frac{120}{100} = 1.2 \) hours.
Distance from \(A\) is the distance covered by the car from \(A\): \( 60 \times 1.2 = 72 \) km.
Check: the car from \(B\) covered \( 40 \times 1.2 = 48 \) km, and \( 72 + 48 = 120 \) km — exactly the initial distance.

Answer: They meet after \( 1.2 \) hours, at \( 72 \) km from \(A\).

Example 4: Average Speed on a Round Trip

Problem: I drove to work at \(40\) km/h and returned along the same route at \(60\) km/h. What was the average speed for the entire trip?

Solution:

Let the one-way distance be \( d \); the total distance is \( 2d \).
Time out: \( \frac{d}{40} \); time back: \( \frac{d}{60} \). Total time: \( \frac{d}{40} + \frac{d}{60} = \frac{3d+2d}{120} = \frac{5d}{120} = \frac{d}{24} \).
Average speed: \( \bar{v} = \frac{2d}{\,d/24\,} = 2d \cdot \frac{24}{d} = 48 \).
Note: the result \(48\) is less than the arithmetic mean \(50\), because more time was spent at the lower speed.

Answer: The average speed is \( 48 \) km/h.

Example 5: Overtaking Problem with Different Departure Times

Problem: A truck leaves at \(50\) km/h. One hour later, a car sets off along the same route at \(75\) km/h. How long after the car's departure does it catch the truck?

Solution:

In the hour before the car departs, the truck covered \( 50 \times 1 = 50 \) km — this is the initial gap.
Both vehicles travel in the same direction, so the gap closes at the difference of speeds: \( 75 - 50 = 25 \) km/h.
Time to close the gap: \( t = \frac{50}{25} = 2 \) hours.
Check: in 2 hours the car covered \( 75 \times 2 = 150 \) km; the truck covered \( 50 + 50 \times 2 = 150 \) km — they are at the same point.

Answer: The car catches the truck \( 2 \) hours after the car's departure.

Common Mistakes

✗ Common mistake: Substituting minutes instead of hours, for example computing \( v = \frac{d}{30} \) when \(30\) is in minutes.

✓ The correct way: Speed in km/h requires time in hours. Convert first: \(30\) minutes \(= 0.5\) hours. Rule of thumb: divide minutes by \(60\) to get hours.

✗ Common mistake: Calculating average speed as a simple arithmetic mean of the speeds, e.g. \( \frac{40+60}{2}=50 \) on a round trip.

✓ The correct way: Average speed = total distance ÷ total time. When the distances are equal but the speeds differ, the result is less than the arithmetic mean (here \(48\), not \(50\)).

✗ Common mistake: Using the difference of speeds for a meeting problem, or the sum for an overtaking problem.

✓ The correct way: Toward each other \(\Rightarrow\) the distance closes quickly, so use the sum of speeds. Same direction (overtaking) \(\Rightarrow\) the gap closes slowly, so use the difference of speeds.

Practice Tips

Tip — Remember the formula triangle: \( v = \frac{d}{t} \). Cover the quantity you want to find and the remaining two show the formula.
Tip — Always check unit consistency before substituting: hours vs. minutes, km vs. metres.
Tip — For two-object problems, draw a number line and mark the starting points and directions; this makes it clear whether it is a meeting or overtaking scenario.
Tip — Sanity check for meeting problems: the sum of the distances covered by both objects must equal the initial distance between them.

Summary and Key Formulas

Fundamental relationship: \( v = \frac{d}{t} \), \( d = v t \), \( t = \frac{d}{v} \).
Meeting (head-on): time \( = \frac{\text{distance}}{v_1 + v_2} \).
Overtaking (same direction): time \( = \frac{\text{gap}}{v_1 - v_2} \).
Average speed \( = \frac{\text{total distance}}{\text{total time}} \) — not the mean of the speeds.
Convert minutes to hours before any calculation in km/h.