8.1 Elapsed Time, Distance, and Speed Word Problems
Key Takeaways
- Elapsed time between two clock times is found by subtraction, borrowing 60 minutes from the hour column whenever the ending minutes are smaller than the starting minutes
- The distance-speed-time relationship distance = speed x time can be rearranged to solve for any one value when the other two are known, but only after converting all time units to match the speed's units (hours, not minutes)
- Minutes convert to a fraction of an hour by dividing by 60, and that converted hour-fraction is what actually gets multiplied by a km/h speed - skipping this conversion is the most common error on these items
- When two vehicles move toward each other, their speeds add into one combined closing speed; when one vehicle chases another in the same direction, only the difference between the two speeds closes the gap
- Because no calculator is allowed, writing down each intermediate result - the converted time, the intermediate product - and rounding to friendly numbers before multiplying saves time under the 35-minute limit
Reading and Subtracting Clock Times
Nearly every Police Problem Solving scenario on the SSPO gives you two clock times and asks how much time separated them - a call received, a unit dispatched, an officer arriving, a shift ending. Elapsed time is always found the same way: subtract the earlier time from the later time, treating hours and minutes as two separate columns rather than one ordinary number.
The complication is that minutes only run from 0 to 59, not from 0 to 99, so a straightforward subtraction sometimes needs to 'borrow' 60 minutes from the hour column exactly the way ordinary subtraction borrows 10 from the next column over.
| Situation | What To Do | Example |
|---|---|---|
| Ending minutes greater than or equal to starting minutes | Subtract minutes directly, then subtract hours directly | 3:50 minus 2:15 gives 1 hour, 35 minutes |
| Ending minutes less than starting minutes | Borrow 60 minutes from the ending hour, add it to the ending minutes, then subtract | 10:15 minus 9:47 becomes 9 hours 75 minutes minus 9 hours 47 minutes, which is 28 minutes |
Work through that second row step by step, because it is the pattern most test-takers stumble on under time pressure. A patrol unit receives a call at 9:47 a.m. and arrives on scene at 10:15 a.m. Because 15 (the ending minutes) is smaller than 47 (the starting minutes), a direct subtraction would go negative - so borrow one hour from the '10' and convert it into 60 extra minutes for the minutes column: 10:15 becomes 9 hours and 75 minutes (60 plus 15). Now subtract normally: 75 minutes minus 47 minutes is 28 minutes, and 9 hours minus 9 hours is 0 hours. The elapsed time is 28 minutes.
A second common variation crosses midnight, where a shift's start time is in the p.m. and its end time is in the a.m. of the next day. The cleanest paper method is to split the calculation into two pieces: the time from the start until midnight, and the time from midnight until the end time, then add those two pieces together. A shift starting at 6:50 p.m. runs 5 hours 10 minutes until midnight (12:00 minus 6:50 is 5:10), then continues 3 hours 10 minutes past midnight to 3:10 a.m. Adding the two pieces, 5:10 plus 3:10 equals 8 hours 20 minutes, gives the total elapsed shift length without ever having to subtract across the calendar boundary directly.
Distance, Speed, and Time: One Formula, Three Uses
Every distance-speed-time question on the SSPO is built from a single relationship: distance = speed x time. Because it is one equation with three quantities, knowing any two lets you solve for the third by rearranging the same formula rather than memorizing three separate rules.
| To Find | Formula |
|---|---|
| Distance | speed x time |
| Speed | distance divided by time |
| Time | distance divided by speed |
The formula only works correctly when the time unit matches the speed unit - a speed given in kilometers per hour must be multiplied by a time expressed in hours, not minutes. Because most word problems give elapsed time in minutes, converting minutes to a fraction (or decimal) of an hour is usually the first real step, done by dividing the minute value by 60.
| Minutes | As a Fraction of an Hour | As a Decimal |
|---|---|---|
| 10 | 1/6 | about 0.167 |
| 15 | 1/4 | 0.25 |
| 20 | 1/3 | about 0.333 |
| 25 | 5/12 | about 0.417 |
| 30 | 1/2 | 0.5 |
| 45 | 3/4 | 0.75 |
Worked example: a patrol vehicle travels at a steady 72 km/h for 25 minutes. First convert 25 minutes to hours: 25/60 equals 5/12 of an hour. Then apply the formula: distance = speed x time = 72 x 5/12. Rather than multiplying by a fraction directly, divide first: 72 divided by 12 is 6, then multiply by 5: 6 x 5 is 30. The vehicle covers 30 km.
A second frequent pattern involves two vehicles moving toward each other (or one catching up to another) rather than a single vehicle covering a single distance. When two objects move toward each other, their speeds add together into one combined 'closing speed' that eats away at the starting gap between them. Two officers on foot patrol start 15 km apart on the same rural road and walk toward each other, one at 6 km/h and the other at 9 km/h. Their combined closing speed is 6 plus 9, or 15 km/h, which exactly matches the starting 15 km gap, meaning they meet after 15 divided by 15, or 1 hour, of walking. If instead one object is chasing another moving in the same direction, the relevant speed is the difference between the two speeds, not the sum - because only the speed advantage closes the gap.
Multi-Step Time-and-Distance Scenarios
The most demanding Police Problem Solving items combine both skills from this section into one scenario: first calculate an elapsed time, then feed that elapsed time into a distance calculation (or vice versa). Solving these cleanly means treating them as two separate, ordinary steps performed one after another - never try to shortcut both steps into a single calculation.
Worked example: a silent alarm at a hardware store is triggered at 11:20 p.m., and the first patrol unit arrives at 11:53 p.m. Step one, find the elapsed time: 53 minus 20 is 33 minutes. Step two, convert that elapsed time to a fraction of an hour: 33/60, which simplifies to 11/20, or as a decimal, 0.55 of an hour. Step three, if investigators want to know the maximum distance a suspect fleeing by bicycle at an average 18 km/h could have covered in that window, multiply: 18 x 0.55. Break the multiplication down for paper work: 18 x 0.5 is 9, and 18 x 0.05 is 0.9, so 9 plus 0.9 is 9.9 km. That 9.9 km figure defines the outer radius of a realistic search area, given the 33-minute head start.
A second version of this multi-step pattern runs the calculation in reverse: given a known distance and a known elapsed time, find the average speed that would have been required to cover it. A witness reports a suspect vehicle passed two fixed checkpoints 14 km apart, and comparing dashcam timestamps shows the vehicle passed the second checkpoint 10 minutes after the first. Converting 10 minutes to hours gives 10/60, which is 1/6 of an hour. Speed = distance divided by time = 14 divided by (1/6). Dividing by a fraction means multiplying by its reciprocal, so 14 x 6 is 84 km/h, the vehicle's average speed between the two checkpoints.
The recurring lesson across every version of these problems is the same: identify which of the three quantities (distance, speed, time) the question is actually asking for, convert every given time value into hours before touching the formula, and perform the conversion and the formula application as two clearly separated steps rather than trying to combine them in your head. Under a 35-minute time limit with no calculator, writing down each intermediate result costs only a few seconds and essentially eliminates arithmetic slips that cost far more time to catch and fix later.
A traffic stop begins at 8:36 a.m. and the officer returns to active patrol at 8:58 a.m. How much time elapsed during the stop?
An overnight shift begins at 10:40 p.m. and ends at 6:15 a.m. the next day. How much total time elapsed?
A cruiser travels at a steady 60 km/h for 45 minutes. How far does it travel?
A speeding vehicle is 3 km ahead of a pursuing cruiser, both traveling in the same direction. The vehicle travels at 78 km/h and the cruiser travels at 90 km/h. How many minutes will it take the cruiser to close the 3 km gap?