# Concept: Time, Speed and Distance

CONTENTS

The basic concept of Time, Speed & Distance is used in solving problems based on Relative motion, circular motion, motion in a straight line, trains etc. Students are advised to understand the concepts of this chapter clearly especially “Concept of Relativity”.

The rate at which a body covers a distance is known as its speed. Mathematically, the above can be represented with the help of the formula :

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$

In other words, speed is the distance covered per unit time.

The basic units to represent speed are km/hr, m/s, etc.

1 km/hr = $\frac{5}{18}$ m/s

1 m/s = $\frac{18}{5}$ km/hr

**Example**: A plane covers the distance of Mumbai to Bhopal in 1 hr and 30 minutes. Calculate the
speed of the plane
if the distance between Bhopal and Mumbai is 800 kms.

**Solution:**

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$

⇒ Speed = $\frac{800}{1.5}$ = 533$\frac{1}{3}$ km/h

**Example**: Express 27 km/h into m/s.

**Solution:**

27 km/h = 27 × $\frac{5}{18}$ m/s = 7.5 m/s

**Example**: Express 12 m/s into km/h.

**Solution:**

12 m/s = 12 × $\frac{18}{5}$ kmph = 43.2 km/h

**Example**: A man travels to his office 50 kms in two hours. He does it partly in a chartered bus
which travels at
the rate of 40 km/h and partly on foot at the rate of 10 km/h. Find how much distance does the man travel on
foot?

**Solution:**

Let x be the distance travelled by bus so the distance travelled on foot = 50
– x

⇒ $\frac{\mathrm{x}}{40}$ + $\frac{\mathrm{(50\; -\; x)}}{10}$ = 2

⇒ x = 40 kms

We have, Distance = Speed × Time

Using the concept of proportionality that we learnt in Variation chapter, we can get some important results.

- When speed is constant, distance travelled is directly proportional to the time travelled.

i.e., D ∝ T

⇒ $\frac{{D}_{1}}{{D}_{2}}$ = $\frac{{T}_{1}}{{T}_{2}}$

- When time travelled is constant, distance travelled is directly proportional to the speed.

i.e., D ∝ S

⇒ $\frac{{D}_{1}}{{D}_{2}}$ = $\frac{{S}_{1}}{{S}_{2}}$

- When distance travelled is constant, speed travelled is inversely proportional to the time travelled.

i.e., S ∝ 1/T

⇒ $\frac{{S}_{1}}{{S}_{2}}$ = $\frac{{T}_{2}}{{T}_{1}}$

So, if S_{1}, S_{2}, S_{3} and T_{1}, T_{2},
T_{3} are the speeds and times for a given body to cover a given distance then it can
be inferred that: S_{1} : S_{2} : S_{3} = $\frac{1}{{T}_{1}}$ : $\frac{1}{{T}_{2}}$ : $\frac{1}{{T}_{3}}$

**Example**: Rashmi walks to her bus stop at 5 km/h and reaches there late by 10 minutes. On the
next day she
increases her speed to 6 km/h and reaches the bus stop 10 minutes early. How far is the bus stop?

**Solution:**

Let x be the distance of the bus stop

The difference in time taken for both speeds = 10 + 10 = 20 minutes = 20/60 hours

⇒ $\frac{\mathrm{x}}{5}$ – $\frac{\mathrm{x}}{6}$ = $\frac{20}{60}$

⇒ x = 10 km/h

Alternate Method:

Ratio of speeds = 5 : 6, hence the ratio of time taken = 6 : 5

So, we can assume that the actual times taken are 6x and 5x respectively.

⇒ 6x – 5x = 20 ⇒ x = 20 minutesHence, time taken at 6 kmph in minutes = 5 × 20 = 100 minutes or 5/3 hours

The distance = 6 × 5/3 = 10 kms

If a particular object/person travels at different speeds during different time-intervals, then average speed can be calculated by dividing the total distance travelled by the total time it takes to travel that distance.

**Example**: Amar travels at a speed of 10 kmph for 2 hours and then increases his speed to 20 kmph
for next 1 hour.
What is his average speed for the entire distance.

**Solution:**

Total distance travelled by Amar = 10 × 2 + 20 × 1 = 40 kms.

Total time taken to cover this distance = 2 + 1 = 3 hours.

∴ Average speed = 40/3 kmph.

Here, average of the two speeds = (10 + 20)/2 = 15 kmph.

**Note:** Average speed is not necessarily equal to average of speeds.

**Example**: A boy covers some distance at the rate of 15 km/h on a bike, and he returns back at a
speed of 12 km/h.
If the time taken for the travel is not known, find the average speed at which the boy travels?

**Solution:**

Let the distance covered by the boy each way b “x”

Average speed = D/T

D = Total distance = 2x,T = Total Time = $\frac{x}{15}$ + $\frac{x}{12}$ (Time = $\frac{\mathrm{Distance}}{\mathrm{Speed}}$)

Average speed = $\frac{\mathrm{2x}}{\mathrm{x/15}}$ + $\frac{x}{12}$ = 13$\frac{1}{3}$ kmph.**Average speed when time travelled is same**

When a person travels at 2 or more than two different speeds for same amount of time, average speed is simple average of individual speeds.

For e.g. if Galib travels at speeds of a, b and c kmph for same amount of time, let us calculate his average speed.

Let the time travelled at each of the three speeds = t hours.

Total time travelled = t + t + t = 3t

Total distance travelled = at + bt + ct

Average speed = $\frac{\mathrm{D}}{\mathrm{T}}$ = $\frac{\mathrm{(at\; +\; bt\; +\; ct)}}{\mathrm{3t}}$ = $\frac{\mathrm{(a\; +\; b\; +\; c)}}{3}$

This is the only case where average speed = average of speeds.

**Average speed when distance travelled is same**

When a person travels at 2 or more than two different speeds for same amount of distance, average speed is same as harmonic mean of individual speeds.

For e.g. if Galib travels at speeds of a, b and c kmph for same amount of distance, let us calculate his average speed.

Let the distance travelled at each of the three speeds = d kms.

Total distance travelled = d + d + d = 3d

Total time travelled = $\frac{\mathrm{d}}{\mathrm{a}}$ + $\frac{\mathrm{d}}{\mathrm{b}}$ + $\frac{\mathrm{d}}{\mathrm{c}}$

Average speed = $\frac{\mathrm{D}}{\mathrm{T}}$ = $\frac{\mathrm{3d}}{\mathrm{(d/a\; +\; d/b\; +\; d/c)}}$ = $\frac{3}{{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}}$

**Note**: When a person travels at two different speeds i.e., x and y, for same amount of distance, the
average speed is also given by the formula $\frac{2x}{{\displaystyle x+y}}$

**Example**: Deepak covers a total distance of 300 kms at three different speeds of 30, 60 and 80
kmph.
Find his average speed for the entire duration.

**Solution:**

Deepak covers 100 kms at each of the three different speeds.

Total time taken = $\frac{100}{30}$ + $\frac{100}{60}$ + $\frac{100}{80}$

∴ Average speed = $\frac{300}{{\displaystyle \frac{100}{30}}+{\displaystyle \frac{100}{60}}+{\displaystyle \frac{100}{80}}}$
= $\frac{3\times 30\times 60\times 80}{30\times 60+60\times 80+80\times 30}$ = 48 kmph.

**Example**: If Karan travels 30 kms at 30 kmph, at what speed should he travel the next 30 kms such
that his
average speed for the entire journey is 48 kmph.

**Solution:**

Total distance travelled by Karan = 30 + 30 = 60 kms

Let speed for second part of the journey be ‘s’ kmph.

∴ Total time taken = $\frac{30}{30}$ + $\frac{30}{\mathrm{s}}$

⇒ Average speed = 48 = $\frac{60}{{\displaystyle \frac{30}{30}}+{\displaystyle \frac{30}{s}}}$

⇒ 4 = 5/(1+30/s)

⇒ 4 + $\frac{120}{\mathrm{s}}$ = 5

⇒ s = 120 kmph

**Example**: Suresh travels by a hovercraft from Vashi to Gateway of India. Every day he catches the
Hovercraft at 9
a.m. One day, however, he is delayed by five minutes because of a railway crossing which he has to cross to
reach Vashi from his home. He normally takes 20 minutes to reach Vashi by his car from Panvel just 30 kms from
Vashi. If the railway crossing is 21 kms from Panvel, find at what speed should. Suresh travel in order to reach
on time?

**Solution:**

The average speed of Suresh S = 20/30 = 2/3 km/h

Suresh must have travelled from Panvel to the railway crossing at the normal speed. The time required by Suresh
to travel the remaining distance from Railway crossing to Vashi is = 2/3 × 9 = 6 minutes, out of these 6
minutes required by Suresh, 5 minutes go waste due to the delay at the crossing. This results in Suresh’s
being required to travel the remaining distance of 9 kms in one hour. Hence, required speed = 9/1 = 9 km/h

Relative speed is the rate of change of distance between two moving objects.

Let us take an example to understand the concept of relative speed.

Suppose two trains are running on parallel tracks in the same direction at 50 kmph. For a
person sitting in one of these train, it would seem that the other train is not moving at
all i.e. it is standing still. In other words, with respect to this person the speed of the other train is zero.

Although the second is actually moving at 50kmph, but for the person sitting in the first
train (which is also moving at 50kmph) it looks stationary.

Let us take one more example. Now suppose you and your friends house are at two ends of a street 500 meters apart. Both of you start from your respective houses to meet each other at speeds of 5 m/s and 10 m/s.

In 1 second, you would travel 5 meters and your friend will travel 10 meters. hence, the distance between you will reduce by 15 meters. Hence, with respect to you, your friend's speed is 15 meters per second.

Relative speed is the most important fundamental used in solving a variety of questions on time and speed. Whenever we consider two bodies in motion simultaneously we must consider the relative speed of the two bodies in order to compute different problems.

- If two bodies are travelling at speeds of 'a' and 'b' (a > b). Their relative speed when they travel in:
- Opposite direction = (a + b) i.e. sum of the speeds
- Same direction = (a - b) i.e. difference of the speeds
- Relative speed of A w.r.t. B is same as relative speed of B w.r.t. A.

**Example**: A policeman running at 12 kmph is trying to catch a thief running at 8 kmph.
If the thief is currently 16 kms ahead of the policeman, how long will it take to catch the thief

**Solution:**

Since both of them are running in same direction, relative speed of policeman
w.r.t. thief = 12 - 8 = 4 kmph

∴ time required to catch the thief = 16/4 = 4 hours.

**Example**: A train travelling at a rate of 25 km/h leaves Bhopal
at 9 am and another train travelling in the same
direction at a speed of 35 km/h starts at 1 pm. How many kms from Bhopal would
both the trains meet and when would they meet?

**Solution:**

The first train travels from 9 am till 1 pm i.e. for 4 hours at 25 kmph.

Distance travelled by it = 4 × 25 = 100 kms.

Now, the second train starts from Bhopal at 35 kmph.

Relative speed = 35 - 25 = 10 kmph

∴ Time taken for the two trains to meet = D/(a - b) = 100/(35 - 25) = 10 hrs

Distance of meeting point from Bhopal will be same as the distance covered by the train starting from Bhopal.

∴ The trains will meet at a distance of 10 × 35 = 350 kms from Bhopal

**Example**: A bus starts from Bhopal at 10 am and goes to Indore.
Another bus follows at a speed of 65 km/h at 11
a.m. Find at what distance from Bhopal will the second bus catch the first bus? (
Given the speed at which the first bus moves is 50 km/h)

**Solution:**

To use relative velocity let us first find out
the distance between both the buses when the
second bus starts moving.

In the first one hour the distance travelled by the first bus is 50 kms.

When the second bus starts moving then, since they are both moving in the same direction, the relative speed is given by : b – a

and time required by the second bus is given by : T = D/(b - a)

⇒ T = 50/(65 - 50) = 10/3 hrs

∴ distance travelled by the second bus is = 10/3 × 65 = 216.45 kms

These questions deal with moving objects (especially trains) with certain length.

- The distance traveled by a train to cross a pole or person is equal to the length of the train.
- The distance traveled by train when it crosses a platform is equal to the sum of the length of the train and length of the platform.
- The distance traveled by train when it crosses the driver of another train is equal to only its own length

**Example**: Two trains 80 m and 120 m in length respectively are running at 45 km/h and 55 km/h.
Find the time they will take to cross each other if it is known that they are travelling in opposite direction?

**Solution:**

The total length to be covered by both the trains together = 120 + 80 = 200 m

Since, the trains are travelling in opposite directions the relative speed will be the addition of the two given speeds = 45 + 55 = 100 km/h = 100 × 5/18 m/s

∴ Time required will be = $\frac{200}{100\times {\displaystyle \frac{5}{18}}}$ = $\frac{36}{5}$ = 7.2 sec.

**Example**: Two trains 80 m and 120 m in length respectively are running at 45 km/h and 55
km/h. Find the time taken by faster train to overtake the slower train?

**Solution:**

The total length covered by both the trains together = 120 + 80 = 200 m

Since, the trains are travelling in same directions the relative speed will be the difference of the two given speeds = 55 - 45 = 10 km/h = 10 × 5/18 m/s

∴ Time required will be = $\frac{200}{10\times {\displaystyle \frac{5}{18}}}$ = $\frac{360}{5}$ = 72 sec.

**Example**: A train 100 m long takes 5 seconds to cross a man walking at the rate of 5 m/s.
Find the speed of the train if the man and the train are travelling in the opposite direction?

**Solution:**

Let the speed of the train be ‘V’ m/s

Relative speed of train and man = V + 5

The time required by the train to cross the man = 5 sec = $\frac{100}{V+5}$

Hence, V = 15 m/s.

**Example**: Two trains of length 100 m and 200 m are travelling in opposite directions
at respective speeds of 30 and 24 kmph. How long will it take for the slower train to cross the driver of the
faster train?

**Solution:**

Relative speed of the two trains = 24 + 30 = 54 kmph = 54 × 5/18 m/s = 15
m/s

Here the slower train has to cross the driver of faster train hence we will not be considering the length of the faster train.

∴ Required time = 200/15 = 40/3 seconds.

**Example**: A train crosses a bridge in 25 seconds when running at 72 km/h and
crosses another bridge twice as long in 20 seconds when running at the rate of 162 km/h. Find the length of the
train and the length of the
bridge?

**Solution:**

72 km/h = 20 m/s and 162 km/h = 45 m/s

Let the length of the train and the smaller bridge be T & B respectively.

$\frac{T+B}{20}$ = 25 and $\frac{T+\mathrm{2B}}{45}$ = 20

Thus, on solving the above equations, we get T and B as 100 m and 400 m respectively.

**Example**: A train 120 m long and travelling at 90 km/h overtakes another
train running at 72 km/h and passes the
second train completely in 50 seconds. Find the length of the second train?

**Solution:**

Let the length of the second train be L then.

Relative speed of the two trains = 90 - 72 = 18 kmph = 5 m/s

⇒ (120 + L)/5 = 50

∴ L = 130 m

**Example**: Two trains, one half as long as the other, cross each other in 20
seconds and the faster one can overtake the slower one in 50 seconds. Find the speed of each of the trains if
the longer train
is 100 m long?

**Solution:**

The length of longer train = 100 m hence, that of shorter train = 50 m.

Let 'a' and 'b' be the speeds of the two trains.

∴ 150/(a + b) = 20 and 150/(a - b)=50.

⇒ a + b = 75 and a – b = 3

On solving the above equations, we get a and b as 39 and 36 m/s respectively.

These problems revolve around the movement of bodies in still and moving fluids. If a boy swims in still water : e.g. a pond or a swimming pool then the speed with which he moves is the speed of that boy in still water. Now imagine the same boy swimming in the river, then he can either swim with the flow of water or against the flow of water. This movement of the boy in the river with the flow of water is called downstream movement whereas the same movement of water against the flow of water is called as an upstream movement. If the speed of the boy is known to be “b” and that of the flow of water is known to be “r” then

Upstream speed (S_{U}) = b - r and Downstream speed (S_{D}) = b + r

- Speed of boat in still water = $\frac{\mathrm{d\; +\; u}}{2}$
- Speed of river = $\frac{\mathrm{d\; -\; u}}{2}$

**Example**: A man can row upstream at the rate of 16 km/h
and downstream at the rate of 20 km/h. Find the speed of
the man in still water and also the speed of the flow of water.

**Solution:**

S_{U} = b - r = 16 and S_{D} = x + y = 20.

On solving the equations we get x = 18 km/h and y = 2 km/h

**Example**: A man can row 6 kms in still water and it
takes him thrice as long as to go up the stream as it takes to
go down the stream. Find the rate of the stream.

**Solution:**

Let the distance to be covered up and
down the stream be D then he the time required by him to go down
the stream and up the stream respectively, are

T_{D} = $\frac{D}{6+y}$
and T_{U} = $\frac{D}{6-y}$

Given, T_{U} = 3 × _{D}

⇒ $\frac{D}{6-y}$ = 3 × $\frac{D}{6+y}$

On solving the given equation we get y = 3

**Example**: A man rows 30 kms upstream and 44 kms downstream in 10 hours on some day and the next
day he rows 40
kms upstream and 55 kms downstream in 13 hours. Find the speed of the man in still water?

**Solution:**

Let the speed of man be ‘b’ km/h and that of the flow of water be
‘r’ then the sum of the upstream and the downstream times must be equal to
the total time required by the man to travel both the distances.

$\frac{30}{\mathrm{u}}$ + $\frac{44}{\mathrm{d}}$ = 10 and $\frac{40}{\mathrm{u}}$ + $\frac{55}{\mathrm{d}}$ = 13

On solving the above two equations we get “d” = 11 km/h and “u” = 5 km/hNow, d = b + r and u = b - r

On solving the above two equations we get b = 8 kmph and r = 3 kmph

**Example**: Amar can swim up to a certain distance with the current in 6 hours and return the same
distance in 9 hours. If the river flows at the rate of 3 km/h then find the speed of Ramesh in still water?

**Solution:**

Let “D” be the distance that Amar swims with the current so he would
have
covered the same distance against the current also. Also let the speed with which Amar can swim be
“x” then we have
the following equations.

⇒ $\frac{\mathrm{D}}{\mathrm{(x\; +\; 3)}}$ = 6 and $\frac{\mathrm{D}}{\mathrm{(x\; -\; 3)}}$ = 9

Thus, x = 15 km/h