Problems Involving Speed

Solving speed problems of more than one step: journeys in stages, two things moving towards each other, one catching up with another, and trains passing posts, tunnels and platforms.

Before you start

Speed is distance per unit of time: speed = distance ÷ time, distance = speed × time and time = distance ÷ speed. Average speed = total distance ÷ total time.

A fraction of an hour is kept as a fraction: $\frac{5}{6}$ h is exactly 50 min, while 0.83 h is only roughly $\frac{5}{6}$ h (49.8 min). 1 h 15 min is $1\frac{1}{4}$ h, not 1.15 h.

To change km/h to m/s, divide by 3.6 (1 km = 1000 m and 1 h = 3600 s); to change m/s to km/h, multiply by 3.6.

Working in Steps

The problems in this lesson take more than one step. None of the steps is new: each one is a speed, a time of day or a unit change you already know. The skill is in seeing which steps are needed and in what order, and in keeping the units right at every step. A journey line, with what is known written on each part, shows the steps.

How to solve a speed problem

1. Write down what is given, with its units, and what is to be found.

2. Draw it: a journey line, with each stage and each mover on it.

3. Put every quantity into units that agree before you multiply or divide (all in metres and seconds, or all in kilometres and hours).

4. Work one step at a time. Keep fractions of an hour exact and round only the final answer.

5. Give the answer in a sentence with its unit, and check that it is sensible.

Example 1: A journey to school in stages

Mei Ling leaves home at 7.20 a.m. She walks 750 m to the MRT station at 1.25 m/s, waits 3 minutes for a train, and then travels 12 km by MRT at an average speed of 48 km/h. At what time does she reach the station near her school?

Solution

Walk: $750 \div 1.25 = 600$ s, and $600 \div 60 = 10$ min.

Wait: 3 min.

MRT: $12 \div 48 = \frac{1}{4}$ h, which is 15 min.

Total time: $10 + 3 + 15 = 28$ min.

7.20 a.m. + 28 min = 7.48 a.m.

Mei Ling reaches the station near her school at 7.48 a.m.

Moving Towards Each Other

Two cars start at the same time from two towns and drive towards each other. Think about the gap between them. In each hour the first car makes the gap shorter by its own distance and the second car makes it shorter by its distance too, so the gap shrinks by the sum of the two distances each hour.

If one car drives at 70 km/h and the other at 50 km/h, then after 1 hour the gap is $70 + 50 = 120$ km shorter. The gap closes at 120 km per hour, and they meet when the whole gap has closed.

Rule: moving towards each other

When two things move towards each other, the gap between them closes by the sum of their speeds in each unit of time.

Time until they meet = the gap at the start ÷ the sum of the speeds. When they move apart in opposite directions, the gap grows by the sum of the speeds instead.

Example 2: Two cars on the same road

Towns A and B are 150 km apart along a straight road. At 9.00 a.m. a car leaves A and drives towards B at a constant speed of 70 km/h. At the same time a van leaves B and drives towards A at a constant speed of 50 km/h.
(a) After how many minutes are they 50 km apart for the first time?
(b) At what time do they meet?

Solution

Each hour the gap closes by $70 + 50 = 120$ km.

(a) To be 50 km apart, the gap must close from 150 km to 50 km, which is $150 - 50 = 100$ km.

Time: $100 \div 120 = \frac{5}{6}$ h, and $\frac{5}{6} \times 60 = 50$ min.

(b) They meet when the whole 150 km has closed: $150 \div 120 = \frac{5}{4} = 1\frac{1}{4}$ h, which is 1 h 15 min.

9.00 a.m. + 1 h 15 min = 10.15 a.m.

Check (b): by then the car has gone $70 \times 1\frac{1}{4} = 87.5$ km and the van $50 \times 1\frac{1}{4} = 62.5$ km, and $87.5 + 62.5 = 150$ km.

(a) They are 50 km apart after 50 minutes. (b) They meet at 10.15 a.m.

Example 3: Where two people meet

Points X and Y are 1150 m apart along a path. Hui Min walks from X towards Y at 1.1 m/s, and at the same moment Ravi walks from Y towards X at 1.2 m/s.
(a) How long is it before they meet?
(b) How far from X do they meet?

Solution

Each second the gap closes by $1.1 + 1.2 = 2.3$ m.

(a) Time: $1150 \div 2.3 = 500$ s. $500 \div 60 = 8$ remainder 20, so 500 s is 8 min 20 s.

(b) In 500 s Hui Min walks $1.1 \times 500 = 550$ m from X.

Check: Ravi walks $1.2 \times 500 = 600$ m, and $550 + 600 = 1150$ m. Hui Min is the slower walker, so the meeting point is nearer X, as 550 m is.

(a) They meet after 8 minutes 20 seconds. (b) They meet 550 m from X.

Catching Up

Now two people travel the same way along the same route, and the faster one starts behind. Each hour the one in front adds its own distance to the gap, and the one behind takes its own distance away from the gap. So the gap shrinks by the difference of the two distances each hour, not the sum.

Rule: catching up in the same direction

First find the gap between them when the one behind sets off.

The gap closes by the difference of the speeds in each unit of time. Time to catch up = the gap ÷ the difference of the speeds.

Example 4: Catching up on the park connector

Siti leaves home at 8.00 a.m. and cycles along a park connector at 12 km/h. Her brother Ali leaves home at 8.30 a.m. and cycles the same way along the same route at 18 km/h. At what time does Ali catch up with Siti, and how far from home are they then?

Solution

The gap when Ali sets off: by 8.30 a.m. Siti has cycled for $\frac{1}{2}$ h, so she is $12 \times \frac{1}{2} = 6$ km ahead.

Each hour Ali cycles 18 km and Siti 12 km, so the gap closes by $18 - 12 = 6$ km.

Time to close 6 km: $6 \div 6 = 1$ h. 8.30 a.m. + 1 h = 9.30 a.m.

Ali has cycled for 1 h, so they are $18 \times 1 = 18$ km from home.

Check: Siti has cycled from 8.00 a.m. to 9.30 a.m., which is $1\frac{1}{2}$ h, and $12 \times 1\frac{1}{2} = 18$ km. They are at the same place.

Ali catches up with Siti at 9.30 a.m., 18 km from home.

Time	Siti's distance from home	Ali's distance from home	Gap
8.30 a.m.	6 km	0 km	6 km
9.00 a.m.	12 km	9 km	3 km
9.30 a.m.	18 km	18 km	0 km

The table shows the gap closing by 3 km every half hour, which is 6 km every hour, the difference of the speeds.

Trains, Tunnels and Platforms

A train is long, so "passing" something takes time even after the front has reached it. Follow one point of the train, its front. A train passes a post (a signal post or a lamp post) from the moment its front reaches the post until its back passes it, so the front travels the length of the train.

A train passes completely through a tunnel (or along a platform) from the moment its front enters until the moment its back leaves. The front travels the length of the tunnel and then the length of the train.

Rule: the train's own length counts

To pass a post, a train travels its own length.

To pass completely through a tunnel or along a platform, a train travels the length of the tunnel or platform + its own length.

Example 5: A post, then a tunnel

A train 120 m long travels at a constant speed. It passes a signal post in 6 s.
(a) Find its speed in m/s and in km/h.
(b) How long does it take to pass completely through a tunnel 480 m long?

Solution

(a) To pass the post, the train travels its own length, 120 m. Speed = $120 \div 6 = 20$ m/s.

In km/h: $20 \times 3.6 = 72$ km/h.

(b) From the moment the front enters to the moment the back leaves, the front travels $480 + 120 = 600$ m.

Time = $600 \div 20 = 30$ s.

(a) The speed is 20 m/s, which is 72 km/h. (b) The train takes 30 seconds to pass through the tunnel.

Example 6: Two tunnels, one train

A train travels at a constant speed. It takes 24 s to pass completely through a tunnel 360 m long, and 16 s to pass completely through a tunnel 216 m long. Find the speed of the train and its length.

Solution

Through either tunnel the train travels the tunnel's length + its own length. The train is the same, so the two distances differ only by the difference of the tunnels.

The longer tunnel is $360 - 216 = 144$ m longer, and the train takes $24 - 16 = 8$ s longer to pass through it.

So the train travels 144 m in 8 s: speed = $144 \div 8 = 18$ m/s.

In 24 s it travels $18 \times 24 = 432$ m. This is the 360 m tunnel + the train, so the train is $432 - 360 = 72$ m long.

Check with the other tunnel: $216 + 72 = 288$ m, and $288 \div 18 = 16$ s.

The train travels at 18 m/s (64.8 km/h) and is 72 m long.

Common Mistakes

Adding the speeds when both go the same way. Moving towards each other, the gap closes by the sum of the speeds; catching up in the same direction, it closes only by the difference.
Forgetting the head start. When one sets off later, first find how far ahead the other already is.
Forgetting the train's own length. Through a 480 m tunnel a 120 m train travels 600 m, not 480 m.
Mixing units. A speed in km/h with a length in metres and a time in seconds gives nonsense. Change everything to metres and seconds, or to kilometres and hours, first.
Writing minutes as a decimal of an hour. 1 h 15 min is $1\frac{1}{4}$ h = 1.25 h, not 1.15 h; and 4.8 min is 4 min 48 s, not 4 min 80 s.
Rounding in the middle. Rounding a time such as $\frac{5}{6}$ h to 0.83 h part-way puts the final answer out. Keep the fraction and round once, at the end.

Practice

Try each question before opening its solution. The questions get harder as you go.

Question 1

Ravi leaves home at 6.50 a.m. He walks 450 m to the bus stop at 1.5 m/s, waits 4 minutes, and then travels 9 km by bus at an average speed of 36 km/h. At what time does he reach school?

ShowHide solution

Walk: $450 \div 1.5 = 300$ s, which is 5 min.

Bus: $9 \div 36 = \frac{1}{4}$ h, which is 15 min.

Total time: $5 + 4 + 15 = 24$ min.

6.50 a.m. + 24 min = 7.14 a.m.

Ravi reaches school at 7.14 a.m.

Question 2

An MRT train travels 2.4 km to the next station in 3 minutes, stops there for 1 minute, and then travels 3.2 km to the following station in 4 minutes. Find its average speed in km/h from leaving the first station to reaching the third.

ShowHide solution

Total distance: $2.4 + 3.2 = 5.6$ km.

Total time, with the stop: $3 + 1 + 4 = 8$ min, which is $\frac{8}{60} = \frac{2}{15}$ h.

Average speed: $5.6 \div \frac{2}{15} = 5.6 \times \frac{15}{2} = 42$ km/h.

Its average speed is 42 km/h.

Question 3

Two cars leave the same point Z at the same time and drive in opposite directions along a straight road, one at 40 km/h and the other at 80 km/h. After how long are they 180 km apart?

ShowHide solution

They drive apart, so each hour the gap grows by $40 + 80 = 120$ km.

Time for the gap to reach 180 km: $180 \div 120 = 1\frac{1}{2}$ h.

Check: $40 \times 1\frac{1}{2} + 80 \times 1\frac{1}{2} = 60 + 120 = 180$ km.

They are 180 km apart after 1 hour 30 minutes.

Question 4

Ali and Siti are at the two ends of a park connector 9 km long. They set off at the same time and cycle towards each other, Ali at 14 km/h and Siti at 16 km/h. After how many minutes do they meet, and how far has Ali cycled by then?

ShowHide solution

Each hour the gap closes by $14 + 16 = 30$ km.

Time: $9 \div 30 = \frac{3}{10}$ h, and $\frac{3}{10} \times 60 = 18$ min.

Ali cycles $14 \times \frac{3}{10} = 4.2$ km.

Check: Siti cycles $16 \times \frac{3}{10} = 4.8$ km, and $4.2 + 4.8 = 9$ km.

They meet after 18 minutes, when Ali has cycled 4.2 km.

Question 5

Two trains are 300 km apart on parallel tracks and travel towards each other at 70 km/h and 80 km/h. How far apart are they 1 hour before they pass each other?

ShowHide solution

The gap closes by $70 + 80 = 150$ km each hour.

In the last hour before they pass, the gap closes by 150 km and ends at 0 km, so 1 hour before they pass the gap is 150 km.

Check: they pass after $300 \div 150 = 2$ h. After 1 h the gap is $300 - 150 = 150$ km.

They are 150 km apart. (The 300 km is not needed.)

Question 6

A bus leaves an interchange at 7.00 a.m. and travels along a main road at 40 km/h. A car leaves the same interchange at 7.15 a.m. and follows the same route at 60 km/h. At what time does the car catch up with the bus, and how far from the interchange?

ShowHide solution

By 7.15 a.m. the bus has travelled $40 \times \frac{1}{4} = 10$ km.

The gap closes by $60 - 40 = 20$ km each hour.

Time: $10 \div 20 = \frac{1}{2}$ h, so the car catches up at 7.15 a.m. + 30 min = 7.45 a.m.

Distance: the car has gone $60 \times \frac{1}{2} = 30$ km.

Check: the bus has travelled for $\frac{3}{4}$ h, and $40 \times \frac{3}{4} = 30$ km.

The car catches up with the bus at 7.45 a.m., 30 km from the interchange.

Question 7

In a race, Siti runs at 3 m/s and starts 60 m ahead of Ravi, who runs at 18 km/h. They start at the same moment. How long does Ravi take to catch up with Siti, and how far has he run by then?

ShowHide solution

Put both speeds in m/s: $18 \div 3.6 = 5$ m/s.

They run the same way, so the gap closes by $5 - 3 = 2$ m each second.

Time to close 60 m: $60 \div 2 = 30$ s.

Ravi runs $5 \times 30 = 150$ m.

Check: Siti runs $3 \times 30 = 90$ m from 60 m ahead, so she is $60 + 90 = 150$ m from Ravi's start, the same place.

Ravi catches up after 30 seconds, having run 150 m.

Question 8

Towns P and Q are 190 km apart. Mr Tan leaves P at 8.00 a.m. and drives towards Q at 80 km/h. Mrs Goh leaves Q at 8.30 a.m. and drives towards P at 70 km/h. At what time do they meet, and how far from P?

ShowHide solution

By 8.30 a.m. Mr Tan has driven $80 \times \frac{1}{2} = 40$ km, so the gap is $190 - 40 = 150$ km.

From 8.30 a.m. they drive towards each other, and the gap closes by $80 + 70 = 150$ km each hour.

Time: $150 \div 150 = 1$ h, so they meet at 9.30 a.m.

Mr Tan has driven for $1\frac{1}{2}$ h: $80 \times 1\frac{1}{2} = 120$ km from P.

Check: Mrs Goh has driven $70 \times 1 = 70$ km, and $120 + 70 = 190$ km.

They meet at 9.30 a.m., 120 km from P.

Question 9

A train 150 m long travels at 54 km/h. How long does it take to pass completely along a platform 210 m long?

ShowHide solution

Speed in m/s: $54 \div 3.6 = 15$ m/s.

Distance from the front reaching the platform to the back leaving it: $210 + 150 = 360$ m.

Time: $360 \div 15 = 24$ s.

The train takes 24 seconds to pass the platform.

Question 10

A train travels at a constant speed. It passes a lamp post in 8 s and passes completely along a platform 300 m long in 23 s. Find the length of the train and its speed in km/h.

ShowHide solution

Passing the post, the train travels its own length in 8 s. Passing the platform, it travels its own length + 300 m in 23 s.

So the extra 300 m takes $23 - 8 = 15$ s, and the speed is $300 \div 15 = 20$ m/s.

Length: in 8 s the train travels $20 \times 8 = 160$ m.

Speed in km/h: $20 \times 3.6 = 72$ km/h.

Check: $(300 + 160) \div 20 = 23$ s.

The train is 160 m long and travels at 72 km/h.

Question 11

On parallel tracks, a train 100 m long travelling at 20 m/s meets a train 140 m long travelling the other way at 10 m/s. How long is it from the moment their fronts meet to the moment their backs separate?

ShowHide solution

When the fronts meet, the back of each train is still behind the meeting point by its own length, so the two backs are $100 + 140 = 240$ m apart along the tracks. The trains have passed each other when the two backs reach each other.

The trains move towards each other, so this gap closes by $20 + 10 = 30$ m each second.

Time: $240 \div 30 = 8$ s.

Check: in 8 s the first train goes $20 \times 8 = 160$ m and the second $10 \times 8 = 80$ m, together 240 m, the two lengths.

It takes 8 seconds.

Key Points

Solve a problem in steps: what is given and what is asked, a journey line, units that agree, one step at a time, an answer in a sentence with its unit.
For a journey in stages, find the time of each stage, add any stops, then add the total time to the time of setting off.
Moving towards each other, the gap closes by the sum of the speeds each unit of time; moving apart, it grows by the sum.
Catching up in the same direction, the gap closes by the difference of the speeds. Find the gap when the second one sets off first.
To pass a post a train travels its own length; to pass through a tunnel or along a platform it travels that length + its own length.
Keep fractions of an hour exact to the end; give values that are not exact to 3 significant figures.