Slope of a time-distance line graph is speed

4.MD.A.25.MD.B.2 · adapt · grade 5

Archetype: Multiplicative Comparison and Unit Rate · step in a 7-type progression

Representative Problem

The line graph shows the relationship between time and distance for Owen riding his bike to a friend's house that is $4$ km from his home. Find how many minutes Owen spent actually moving on the bike.

(Figure) A line graph titled "Distance from Home." The horizontal axis is time: the major mark $1$ stands for $1$ hour, and $1$ hour is divided into $4$ small squares, so each small square is $15$ minutes. The vertical axis is distance from home in kilometers (km): $0$ , $1$ , $2$ , $3$ , $4$ . Starting from $(0$ min, $0$ km $)$ , the line rises from $0$ km to $2$ km over the first $1$ square, then stays flat at $2$ km for $2$ squares, then rises from $2$ km to $3$ km over $1$ square, then stays flat at $3$ km for $1$ square, and finally rises from $3$ km to $4$ km over the last $1$ square. (Sloped segments are where Owen was riding; flat segments are where he had stopped.)

(Note: distances are kept in kilometers because the answer is read directly off this metric time-distance graph; the figure has not been redrawn in imperial units.)

Show solution

Understand

A time-distance line graph shows Owen biking to a friend's house 4 km away. Each small square is 15 minutes. The line rises 0 to 2 km over 1 square, stays flat at 2 km for 2 squares, rises 2 to 3 km over 1 square, stays flat at 3 km for 1 square, then rises 3 to 4 km over 1 square. Sloped parts are riding; flat parts are stopped. I must find how many minutes he spent actually moving.

Givens

Each small grid square = 15 minutes
Sloped (rising) segments mean Owen is moving; flat segments mean he is stopped
Rising segments: 0->2 km (1 square), 2->3 km (1 square), 3->4 km (1 square)
Flat segments: at 2 km for 2 squares, at 3 km for 1 square

Unknowns

Total minutes Owen spent moving on the bike

Constraints

Only the sloped segments count as moving time

Plan

#1 Draw a Diagram · also uses: #8 Analyze the Units

I read the graph to separate sloped (moving) squares from flat (stopped) squares, count only the sloped squares, then convert squares to minutes using 1 square = 15 minutes.

Execute

#1 Draw a Diagram 5.MD.B.2

The sloped (rising) parts are: 0 to 2 km over 1 square, 2 to 3 km over 1 square, and 3 to 4 km over 1 square. That is 1 + 1 + 1 = 3 squares of moving. The flat parts (2 + 1 = 3 squares) are stops and do not count.

1 + 1 + 1 = 3 \text{ squares moving}

A rising line means distance is changing, so he is riding only there.

#8 Analyze the Units 4.MD.A.2

Each square is 15 minutes, so 3 moving squares = 3 x 15 = 45 minutes.

3 \times 15 = 45 \text{ minutes}

Multiplying square-count by minutes-per-square gives total moving time in the right unit.

Answer: 45 minutes

Review

There are 6 squares of graph time total (3 sloped + 3 flat) = 90 minutes for the whole trip; the moving half is 3 squares = 45 minutes, which is less than the full trip and sensible. The 4 km covered in 45 minutes of riding is a believable bike pace.

Work it as a subproblem total (tool 7): whole trip is 90 minutes, stopped time is the 3 flat squares = 45 minutes, so moving = 90 - 45 = 45 minutes - the same answer.

Standards · min grade 5

5.MD.B.2 Make a line plot to display a data set and solve problems using the data — Reading the graph to tell moving (sloped) from stopped (flat) segments
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money — Converting grid squares to minutes of moving time

💡 On a time-distance graph the slanted parts are when you're moving - count those squares and turn them into minutes!