The distance is a physical concept that describes how far or how near is an object from a point. It also represents length of space between two points. This is a scalar quantity.

The speed is said to be a measure of distance per unit time, i.e. it is the distance covered in a unit time interval.

Speed and distance are two quantities that are directly proportional to each other. The third quantity introduced to make them equivalent is the time factor. Whenever we talk about speed and distance, the time factor can never be ignored.

## Speed & Distance Definition

Speed is the rate at which the object is moving for example 40 km/h which means the object is covering a path of 40 kilometers in an hour. Distance is the path that is covered by a moving object where the object can be a vehicle, a person or any other thing which is moveable.

The distance traveled by an object is always directly proportional to the speed with which that object is moving. This implies, if the speed of the object increases the distance will also increase accordingly and if the speed of the object will decrease, the distance will also decrease in a certain period of time.

The factor time that is introduced to make an equivalence relation between speed and distance is also a factor of distance which is directly proportional to it and hence, inversely proportional to speed.

## Speed & Distance Formula

The formula that is used to determine the distance, speed or time when the any two of the three values are known is given below:

$Distance = Speed \times Time$

That is, distance traveled by an object is the product of the speed of that object and the time taken by it to cover the distance.
From the above formula we can also deduce that,

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

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

## How to Solve Speed & Distance

For finding out any value from the formulas mentioned above, the give values should be in the same type of format, that are in the same units. For example if we are given speed in km/h that is kilometers per hour and distance is given in m (meters), then we must either convert speed to m/s that is meters per second or else convert the distance in kilometers. In short we mean, the units must always be same for any computation to be performed for all the factors, speed, distance and time.

If we want to convert the distance, speed or time in various units we can use the following conversions

1 km = 1000 m = 100,000 cm

1 km/h = $\frac{1000 m}{(60 \times 60)s}$ = $\frac{5}{18}$ m/s

1 h = 60 min = 60 $\times$ 60 s = 3600 s

This is a distance versus time graph. It is of no direct reference here. But we are discussing it as this graph tells the concept of the term speed. The slope of the distance versus time curve gives us the value of the speed of the object. If we have a straight line curve, this implies the object is moving with uniform speed. If we have a curved graph, then we either talk about average speed of the object by finding the average of the speeds of the object at various points of time or else we talk about speed of the object at a particular instant.

## Speed & Distance Problems

Let us see some examples based on distance and speed for an understanding the concept in a better manner.

Example 1:

What is the speed of the man in running up 100 m in 10 s. Express in km/h?

Solution:

The distance covered by the man = 100 m.

The time taken by the man = 10 s

Therefore, speed required by the $man$ = $\frac{distance}{time}$ = $\frac{100}{10}$ = $10m/s$

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

$\Rightarrow$ $10 m/s$ = $\frac{180}{5}$ km/h = $36 km/h$

The speed required by the man to run up 100 m distance in 10 s is 36 km/h.
Example 2:

A train is passing a pole in exactly 11 seconds and the same train passes the platform completely in 44 seconds. If the length of the platform is given to be 600 m, then find the length and the speed of the train.

Solution:

Let the length of the train be y m.

The distance covered by train in passing the platform is the same as the sum of the length of train and the length of platform.

$\Rightarrow$ Distance covered in passing platform = y + 600

According to question we have,

While passing the pole, speed of the $train$ = $\frac{y}{11}$m/s

Again, while passing the platform, speed of the train = $\frac{(y + 600)}{44}$m/s

Therefore, $\frac{y}{11}$ = $\frac{(y + 600)}{44}$

$\Rightarrow$ $y$ = $\frac{(y + 600)}{4}$

$\Rightarrow$ $4y$ = $y + 600$

$\Rightarrow$ $3y$ = $600$

$\Rightarrow$ $y$ = $200$

Therefore, the length of the train is 200 meters.

Also, the speed of the train is $\frac{200}{11}$ = 18.18 m/s.