Speed, distance and time

How long will the bus take to reach Vijayawada? How fast must you cycle to reach school on time? Ques­tions like these need three ideas: dis­tance, time and speed. They are joined by one sim­ple for­mula. In this les­son you will learn that for­mula, how to change between km/h and m/s, what aver­age speed really means, and how speed and time com­pare when the dis­tance is the same.

Dis­tance, time and speed#

Dis­tance. The length of the path trav­elled by an object or a per­son between two places.

Time. The dura­tion, in hours, min­utes or sec­onds, spent cov­er­ing a dis­tance.

Speed. The dis­tance cov­ered in one unit of time.

If the speed of a car is 4 km/h (kilo­me­tres per hour), it cov­ers 4 km in 1 hour.

The for­mula#

$$\displaystyle \text{speed} = \frac{\text{distance}}{\text{time}}$$

Turn it round to get the other two:

$$\displaystyle \text{distance} = \text{speed} \times \text{time} \qquad \text{time} = \frac{\text{distance}}{\text{speed}}$$

Learn the first one well. The other two come from it.

Exam­ple 1#

A car trav­els at 85 km/h and reaches its des­ti­na­tion in 5 hours. How far did it go?

$$\text{distance} = \text{speed} \times \text{time} = 85 \times 5 = 425 \text{ km}$$

Exam­ple 2#

Abhishek rides his bike at 150 km/h. How far does he go in 3 hours?

In 1 hour he cov­ers 150 km. In 3 hours he cov­ers $150 \times 3 = 450$ km.

Or by the for­mula: dis­tance $= 150 \times 3 = 450$ km.

Exam­ple 3#

A cycle cov­ers 75 km in 5 hours. How far does it go in 6 hours, at the same speed?

The speed is $75 \div 5 = 15$ km/h. In 6 hours it goes $15 \times 6 = 90$ km.

Or by pro­por­tion: $\displaystyle \frac{6}{5} \times 75 = 90$ km.

Chang­ing km/h into m/s#

1 km = 1000 m and 1 hour = 60 × 60 = 3600 sec­onds. So

$$\displaystyle 1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}$$

and the other way round,

$$\displaystyle 1 \text{ m/s} = \frac{18}{5} \text{ km/h}$$

• To change km/h into m/s, mul­ti­ply by $\displaystyle \frac{5}{18}$.
• To change m/s into km/h, mul­ti­ply by $\displaystyle \frac{18}{5}$.

Exam­ple 4#

Change 18 km/h into m/s.

$$\displaystyle 18 \times \frac{5}{18} = 5, \text{ so } 18 \text{ km/h is } 5 \text{ m/s}$$

Exam­ple 5#

Change 10 m/s into km/h, and 15 m/s into km/h.

$$\displaystyle 10 \times \frac{18}{5} = 36 \qquad 15 \times \frac{18}{5} = 54$$

So 10 m/s is 36 km/h, and 15 m/s is 54 km/h.

Exam­ple 6: The speed of sound#

A gun is fired 3.32 km away from Ajay. He hears the sound 10 sec­onds later. Find the speed of sound in m/s.

3.32 km = 3320 m.

$$\displaystyle \text{speed} = \frac{3320}{10} = 332 \text{ m/s}$$

Aver­age speed#

Aver­age speed.

$$\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}$$

It is not the aver­age of the speeds. Here is why.

Going and com­ing back#

A car goes from A to B at speed $s_1$ and comes back from B to A at speed $s_2$. Let the dis­tance from A to B be $d$.

Time going $\displaystyle = \frac{d}{s_1}$ and time com­ing back $\displaystyle = \frac{d}{s_2}$.

$$\displaystyle \begin{aligned} \text{average speed} &= \frac{d + d}{\frac{d}{s_1} + \frac{d}{s_2}} \\ &= \frac{2d}{d\left(\frac{1}{s_1} + \frac{1}{s_2}\right)} \\ &= \frac{2}{\frac{s_1 + s_2}{s_1 s_2}} \\ &= \frac{2s_1 s_2}{s_1 + s_2} \end{aligned}$$

For two equal dis­tances at speeds $s_1$ and $s_2$, the aver­age speed is $\displaystyle \frac{2s_1 s_2}{s_1 + s_2}$.

The same work­ing for three equal dis­tances at speeds $s_1$, $s_2$ and $s_3$ gives

$$\displaystyle \text{average speed} = \frac{3d}{\frac{d}{s_1} + \frac{d}{s_2} + \frac{d}{s_3}} = \frac{3s_1 s_2 s_3}{s_1 s_2 + s_2 s_3 + s_3 s_1}$$

Exam­ple 7#

Deepak dri­ves some­where by car at 25 km/h and comes back on a scooter at 15 km/h. Find his aver­age speed for the whole jour­ney.

$$\displaystyle \text{average speed} = \frac{2 \times 25 \times 15}{25 + 15} = \frac{750}{40} = 18.75 \text{ km/h}$$

The plain aver­age of 25 and 15 is 20, but the answer is less than that. He spends more time at the slower speed, so the slow part counts for more.

Exam­ple 8#

Udai trav­els half his jour­ney by train at 120 km/h and the other half by car at 80 km/h. Find his aver­age speed.

The two halves are equal dis­tances, so

$$\displaystyle \text{average speed} = \frac{2 \times 120 \times 80}{120 + 80} = \frac{19200}{200} = 96 \text{ km/h}$$

Speed and time for the same dis­tance#

If two peo­ple cover the same dis­tance $d$, one at speed $s_1$ in time $T_1$ and the other at speed $s_2$ in time $T_2$, then

$$\displaystyle s_1 = \frac{d}{T_1} \qquad s_2 = \frac{d}{T_2}$$

Divide one by the other. The $d$ can­cels:

$$\displaystyle \frac{s_1}{s_2} = \frac{T_2}{T_1} \qquad \text{so} \qquad s_1 : s_2 = T_2 : T_1$$

For a fixed dis­tance, speed and time are in inverse pro­por­tion. The faster per­son takes less time.

Exam­ple 9#

Karan's speed is 40 km/h and Arjun's speed is 60 km/h. Find the ratio of the times they take to cover the same dis­tance.

$$T_1 : T_2 = s_2 : s_1 = 60 : 40 = 3 : 2$$

Karan, the slower one, takes longer.

Exam­ple 10#

A car takes half the time that a truck takes to go from Luc­know to Mum­bai. The truck takes 20 hours. The car's speed is 120 km/h. Find the speed of the truck.

The car takes 10 hours and the truck takes 20 hours, so

$$\displaystyle \frac{s_{\text{car}}}{s_{\text{truck}}} = \frac{T_{\text{truck}}}{T_{\text{car}}} = \frac{20}{10} = \frac{2}{1}$$

The car is twice as fast, so the truck's speed is $120 \div 2 = 60$ km/h.

Late and early#

Exam­ple 11#

If a man trav­els at 30 km/h, he reaches his des­ti­na­tion 10 min­utes late. If he trav­els at 42 km/h, he reaches 10 min­utes early. Find the dis­tance.

Let the cor­rect time be $T$ hours. 10 min­utes is $\displaystyle \frac{10}{60} = \frac{1}{6}$ hour.

At 30 km/h he takes $\displaystyle T + \frac{1}{6}$ hours. At 42 km/h he takes $\displaystyle T - \frac{1}{6}$ hours. The dis­tance is the same both times:

$$\displaystyle \begin{aligned} 30\left(T + \frac{1}{6}\right) &= 42\left(T - \frac{1}{6}\right) \\ 30T + 5 &= 42T - 7 \\ 5 + 7 &= 42T - 30T \\ 12 &= 12T \\ T &= 1 \end{aligned}$$

$$\displaystyle \text{distance} = 30 \times \left(1 + \frac{1}{6}\right) = 30 \times \frac{7}{6} = 35 \text{ km}$$

Check with the other speed: $\displaystyle 42 \times \frac{5}{6} = 35$ km.

Prac­tice#

1. A bus trav­els at 60 km/h for 4 hours. How far does it go?
2. A train cov­ers 270 km in 3 hours. Find its speed.
3. How long does a car take to cover 210 km at 70 km/h?
4. Change 72 km/h into m/s.
5. Change 25 m/s into km/h.
6. A man walks to a tem­ple at 4 km/h and comes back at 6 km/h. Find his aver­age speed.
7. A car cov­ers three equal dis­tances at 20 km/h, 30 km/h and 60 km/h. Find its aver­age speed.
8. The speeds of two cars are in the ratio $5 : 4$. Find the ratio of the times they take for the same jour­ney.
9. A boy walks to school at 4 km/h and reaches 5 min­utes late. If he walks at 5 km/h, he reaches 10 min­utes early. How far is his school?

Answers#