DISTANCE, TIME AND SPEED

**Category : **6th Class

**Learning Objective**

__ __

To find overage speed of a vertical.

To find time to cover a given distance.

**TIME AND DISTANCE**

Motion / Movement occurs when a body of any shape and size changes its position with respect to any external stationary point.

The mathematical equation that describe the motion has three variables Speed, Time and Distance, which are connected by the following formula

Distance = Speed x Time

From the above equation, we can have the following conclusions:

(a) If speed is constant then distance and time are directly proportional to each other, i.e.

Distance \[\propto \] Time.

(b) If time is constant, then distance and speed are directly proportional to each other i.e

Distance \[\propto \] Speed.

(c) When distance is constant then speed and time are inversely proportional to each other i.e.

Speed c\[\propto \,\frac{1}{Time}\]

Normally speed is measured in km/hr or m/s.

\[1\,km/hr=\frac{1000\,m}{3600\,s}=\frac{5}{18}\,m/s\]

or \[1\,m/s\,=\frac{18}{5}km/hr\]

** **

**AVERAGE SPEED:**

It is defined as the ratio of total distance covered to the total time taken by an object. If an object travels \[{{d}_{1}},{{d}_{2}},\,{{d}_{3}},\,....{{d}_{n}}\] metres with different speeds \[{{s}_{1}},\,{{s}_{2}},\,{{s}_{3}},....{{s}_{n}}\] metres/ sec in time \[{{t}_{1}},\,{{t}_{2}},\,{{t}_{3}},......{{t}_{n}}\] seconds respectively, then average speed \[{{S}_{a}}\] is given by

\[{{S}_{a}}=\frac{\text{Total}\,\text{Distance}\,\text{Travelled}}{\text{Total}\,\text{Time}\,\text{Taken}}\]

A car can cover 350 km in 4 hours. If its speed is decreased by \[12\frac{1}{2}\] kmph, how much time does the car take to cover a distance of 450 km?

**Solution:**

\[\text{Speed}\,\text{=}\frac{\text{Distance}}{\text{Time}}\,=\frac{350}{4}=87\frac{1}{2}\,\text{kmph}\]

Now this is reduced by \[12\frac{1}{2}\] kmph.

Hence, speed is 75 kmph.

Travelling at this speed, the time taken == 450/75 = 6 hours.

*play_arrow*DISTANCE, TIME AND SPEED

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