Skip to content
Algebra · Chapter 20

Speed, Distance, and Time

Distance equals speed times time, and almost everything else follows: average speed (it is not the average of the speeds!), meeting and catching up, and walking on escalators and moving walkways.

You'll need
open on YouTube
20.0a0%

Distance, Speed and Average Speed

In this section you will be able to use distance = speed × time, find an average speed the right way, and spot the average-speed trap.

You hike 4 miles out at 4 mph and come back the same way at 2 mph. Your average speed is not 3 mph. Why not?

Every problem in this chapter rests on one formula: distance = speed × time. Know two of the three and you get the third.

Average speed is total distance divided by total time. Look: on the way back you are slow, so you spend longer at the slow speed. That drags the average below the plain mean of the speeds.

Exploreplay first, formulas later

4 mi at 4 mph, then 4 mi at 2 mph: 8 mi in 3 h, average speed 8/3 mph (not the mean of the speeds, 3)

Your guess3 mph
Out speed4 mph
Back speed2 mph
Distance (equal shares of the road)4 mi4 miTime (how long each part lasts)1 h2 hclock: 0 of 3 h
  • Out: 4 mi at 4 mph
  • Back: 4 mi at 2 mph
Average speed = ? Make a guess!

Average speed=total distancetotal time\text{Average speed} = \frac{\text{total distance}}{\text{total time}}

Your guess: 3 mph. The legs take 1 and 2 h.

Name it

Theorem 20.0.1 · Distance, Speed and Time

Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}

Equivalently,

Speed=DistanceTimeTime=DistanceSpeed\text{Speed} = \frac{\text{Distance}}{\text{Time}} \qquad \text{Time} = \frac{\text{Distance}}{\text{Speed}}

Keep the units matched: miles per hour goes with time in hours. Convert minutes first.

Joe starts at 0 mi at 60 mph: 40 mi in 0.67 h

Clock0 h (0 min)
Joe's speed60 mph
010203040Joe
00.250.50.751010203040time (h)position (mi)
t = 0 h (0 min)

distance=speed×time\text{distance} = \text{speed} \times \text{time}: a steeper line on the graph means a faster mover.

t = 0 h (0 min): Joe at 0 mi

Theorem 20.0.2 · Average Speed

Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

Watch itpredict, then reveal

Example 20.1

AMC 10

Henry decides one morning to do a workout, and he walks 34\frac{3}{4} of the way from his home to his gym. The gym is 22 kilometers away from Henry's home. At that point, he changes his mind and walks 34\frac{3}{4} of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks 34\frac{3}{4} of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked 34\frac{3}{4} of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point AA kilometers from home and a point BB kilometers from home. What is ∣A−B∣|A - B|?

Geometric growth (its total is revealed at the end of the example)

r = ?
Running total after each term
S∞ = 8/15

Sum 8/15; sum of the squares 64/375

  1. 1

    Put home at 00 and the gym at 22. Where does Henry first turn around?

    Predict first:
open on YouTube

Try it

Problem 20.1.1

AMC 10

A cart rolls down a hill, traveling 55 inches the first second and accelerating so that during each successive 11-second time interval, it travels 77 inches more than during the previous 11-second interval. The cart takes 3030 seconds to reach the bottom of the hill. How far, in inches, does it travel?

Problem 20.1.4(additional)

AMC 10

Sam drove 9696 miles in 9090 minutes. His average speed during the first 3030 minutes was 6060 mph (miles per hour), and his average speed during the second 3030 minutes was 6565 mph. What was his average speed, in mph, during the last 3030 minutes?

1/2 h at 60 mph (30 mi), then 1/2 h at 65 mph (32.5 mi), then 1/2 h at 60 mph (30 mi): 92.5 mi in 3/2 h, average speed 185/3 mph (not the mean of the speeds, 185/3)

Your guess62 mph
First 30 min speed60 mph
Second 30 min speed65 mph
Last 30 min speed60 mph
Distance (speed × time)30 mi32.5 mi30 miTime (each part lasts the same)1/2 h1/2 h1/2 hclock: 0 of 3/2 h
  • First 30 min: 30 mi at 60 mph
  • Second 30 min: 32.5 mi at 65 mph
  • Last 30 min: 30 mi at 60 mph

Total so far: 92.5 mi in 3/2 h

Average speed = ? Make a guess!

Average speed=total distancetotal time\text{Average speed} = \frac{\text{total distance}}{\text{total time}}

Your guess: 62 mph. The legs take 1/2 and 1/2 and 1/2 h. Total so far: 92.5 mi in 3/2 h.

Problem 20.1.6(additional)

AMC 10

Samia set off on her bicycle to visit her friend, traveling at an average speed of 1717 kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at 55 kilometers per hour. In all it took her 4444 minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?

Contest cornerreal contest problems

Problem 20.1.2

AMC 10

Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at 44 miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to 22 miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at 33 miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?

6 mi at 4 mph, then 6 mi at 2 mph, then 6 mi at 3 mph: 18 mi in 13/2 h, average speed 36/13 mph (not the mean of the speeds, 3)

Your guess3 mph
Gentle half, up speed4 mph
Steep half, up speed2 mph
Steep half, down speed3 mph
Distance (equal shares of the road)6 mi6 mi6 miTime (how long each part lasts)3/2 h3 h2 hclock: 0 of 13/2 h
  • Gentle half, up: 6 mi at 4 mph
  • Steep half, up: 6 mi at 2 mph
  • Steep half, down: 6 mi at 3 mph
Average speed = ? Make a guess!

Average speed=total distancetotal time\text{Average speed} = \frac{\text{total distance}}{\text{total time}}

Your guess: 3 mph. The legs take 3/2 and 3 and 2 h.

Problem 20.1.5(additional)

AIME

Every morning, Aya does a 99 kilometer walk, and then finishes at the coffee shop. One day, she walks at ss kilometers per hour, and the walk takes 44 hours, including tt minutes at the coffee shop. Another morning, she walks at s+2s + 2 kilometers per hour, and the walk takes 22 hours and 2424 minutes, including tt minutes at the coffee shop. This morning, if she walks at s+12s + \frac{1}{2} kilometers per hour, how many minutes will the walk take, including the tt minutes at the coffee shop?

Problem 20.1.7(additional)

AMC 10

Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 4040 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.020.02 gallons per mile. On the whole trip he averaged 5555 miles per gallon. How long was the trip in miles?

Problem 20.1.8(additional)

AIME

Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the 5050-mile mark at exactly the same time. How many minutes has it taken them?

Checkpoint for section 20.0a

Get 2 of 3 right to master this section. One try each, no hints.
1

A cyclist rides 3030 km at 1515 km/h and then another 3030 km at 1010 km/h. What is her average speed for the whole ride, in km/h?

2

A train covers 150150 miles in 22 hours and 3030 minutes. What is its average speed in miles per hour?

3

You drive for 11 hour at 4040 mph and then for 11 hour at 6060 mph. What is your average speed, in mph?

20.0b0%

Relative Motion

In this section you will be able to add or subtract speeds for movers heading toward or after each other, and handle escalators and moving walkways.

Walking down an up-escalator, Al counts 150 steps. Bob walks up and counts only 75. How many steps does the escalator show?

When two things move, look at the gap between them. Heading toward each other, the gap closes at the sum of the speeds. Chasing, it closes at the difference.

A moving walkway or escalator just adds its speed to everyone riding it. The steps you count are only the ones your own feet take.

Exploreplay first, formulas later

Walkway from 0 to 100 ft, moving at 6 ft/s. Al starts at 0 ft at 0 ft/s: 100 ft in 16.67 s; Bob starts at 0 ft at 4 ft/s: 120 ft in 17 s; Cy starts at 0 ft at 6 ft/s: 120 ft in 24 s. Al is exactly halfway between the others at t = 11 s, 66 ft from the start

Clock0 s
Al's speed0 ft/s
Bob's speed4 ft/s
Cy's speed6 ft/s
Walkway →0306090120AlBobCy
051015200306090120time (s)position (ft)
t = 0 s

distance=speed×time\text{distance} = \text{speed} \times \text{time}: a steeper line on the graph means a faster mover.

t = 0 s: Al at 0 ft, Bob at 0 ft, Cy at 0 ft

Watch itpredict, then reveal

Example 20.2

AIME

Al walks down to the bottom of an escalator that is moving up and he counts 150150 steps. His friend, Bob, walks up to the top of the escalator and counts 7575 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)

Race track (its total is revealed at the end of the example)

Clock0 s
Escalator →0????AlBob
0204060800????time (s)position (steps)
t = 0 s
  • Al walked 0 steps · the belt moved him ? (hidden until the end of the example) steps
  • Bob walked 0 steps · the belt moved him ? (hidden until the end of the example) steps

distance=speed×time\text{distance} = \text{speed} \times \text{time}: a steeper line on the graph means a faster mover.

t = 0 s: Al at ? steps, Bob at ? steps. Al walked 0 steps, Bob walked 0 steps

  1. 1

    Only the ratio of speeds matters, so say Bob walks 11 step per second and Al walks 33. Al counts 150150 steps. How many seconds is he on the escalator?

    Predict first:
open on YouTube

Example 20.3

AIME

A 100100 foot long moving walkway moves at a constant rate of 66 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 44 feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 88 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person.

Race track (its total is revealed at the end of the example)

Clock0 s
Walkway →0306090120AlBobCy
051015200306090120time (s)position (ft)
t = 0 s

distance=speed×time\text{distance} = \text{speed} \times \text{time}: a steeper line on the graph means a faster mover.

t = 0 s: Al at 0 ft, Bob at 0 ft, Cy at 0 ft

  1. 1

    How fast does Bob move over the ground?

    Predict first:
open on YouTube

Try it

Problem 20.1.3

MATHCOUNTS

A car passes point AA driving at a constant rate of 6060 km per hour. A second car, traveling at a constant rate of 7575 km per hour, passes the same point AA a while later and then follows the first car. It catches the first car after traveling a distance of 7575 km past point AA. How many minutes after the first car passed point AA did the second car pass point AA?

Car 1 starts at 0 km at 60 km/h: 100 km in 1.67 h; Car 2 starts at 0 km at 75 km/h: 100 km in 1.33 h. Car 1 and Car 2 do not meet by t = 1.5 h

Clock0 h (0 min)
Car 2 starts at0 h (0 min)
0255075100Car 1Car 2
00.380.751.131.50255075100time (h)position (km)
t = 0 h (0 min)

distance=speed×time\text{distance} = \text{speed} \times \text{time}: a steeper line on the graph means a faster mover.

t = 0 h (0 min): Car 1 at 0 km, Car 2 at 0 km

Contest cornerreal contest problems

Problem 20.1.9(additional)

AMC 12

Andrea and Lauren are 2020 kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of 11 kilometer per minute. After 55 minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?

Andrea starts at 0 km at 0.75 km/min: 3.75 km in 5 min; Lauren starts at 20 km at 0.25 km/min: 20 km in 80 min. Andrea and Lauren do not meet by t = 30 min

Clock0 min
05101520AndreaLauren
07.51522.53005101520time (min)position (km)
t = 0 min

distance=speed×time\text{distance} = \text{speed} \times \text{time}: a steeper line on the graph means a faster mover.

t = 0 min: Andrea at 0 km, Lauren at 20 km

Problem 20.1.10(additional)

AMC 12

While Steve and LeRoy are fishing 11 mile from shore, their boat springs a leak, and water comes in at a constant rate of 1010 gallons per minute. The boat will sink if it takes in more than 3030 gallons of water. Steve starts rowing towards the shore at a constant rate of 44 miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?

Problem 20.1.13(additional)

AIME

Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at 2020 miles per hour, and Steve rides west at 2020 miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly 11 minute to go past Jon. The westbound train takes 1010 times as long as the eastbound train to go past Steve. The length of each train is mn\frac{m}{n} miles, where mm and nn are relatively prime positive integers. Find m+nm + n.

Checkpoint for section 20.0b

Get 2 of 3 right to master this section. One try each, no hints.
1

Two trains 300300 km apart head toward each other at 7070 km/h and 8080 km/h. After how many hours do they meet?

2

A runner leaves a gate at 66 mph. Thirty minutes later a cyclist leaves the same gate on the same path at 1818 mph. How many minutes after the cyclist starts does she catch the runner?

3

A 7070 m moving walkway moves at 22 m/s. You walk on it, in its direction, at 1.51.5 m/s. How many seconds does it take you to reach the end?

20.1

Practice Problems

In this section you will be able to put rates, equations and geometry together on two harder AIME problems.

Two last AIME problems: one hides a system of equations with whole-number answers, the other sends three bikers around a field. Write distance = speed × time for every person first.

The other practice problems sit in sections 20.0a and 20.0b, next to the ideas they use. Want them all on one page? Open the problem set.

Contest cornerreal contest problems

Problem 20.1.11(additional)

AIME

Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers 7474 kilometers after biking for 22 hours, jogging for 33 hours, and swimming for 44 hours, while Sue covers 9191 kilometers after jogging for 22 hours, swimming for 33 hours, and biking for 44 hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.

Problem 20.1.12(additional)

AIME

Ana, Bob, and Cao bike at constant rates of 8.68.6 meters per second, 6.26.2 meters per second, and 55 meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point DD on the south edge of the field. Cao arrives at point DD at the same time that Ana and Bob arrive at DD for the first time. The ratio of the field's length to the field's width to the distance from point DD to the southeast corner of the field can be represented as p:q:rp : q : r, where pp, qq, and rr are positive integers with pp and qq relatively prime. Find p+q+rp + q + r.