Basics of Speed, time and distance and its applications

Speed, Time and Distance and Its Applications

Concept of speed, time and distance is based on the formula
Speed × time = distance

Speed (s):
Speed can be defined as rate at which distance is covered during the motion. It is measured in terms of distance per unit time. (i.e. m/s, km/hour, m/min, km/min, etc.).
Speed = $$\frac{{{\text{Distance}}}}{{{\text{Time}}}}$$

Time (t):
It is the time duration over which the movement has occurred. The unit used for measuring time is synchronous with denominator of the unit used for measuring speed. Thus, if the speed is measured in terms of km/h then time is measured in hours.
Time = $$\frac{{{\text{Distance}}}}{{{\text{Speed}}}}$$

Distance (d):
It is the displacement of the body during the motion.
Distance= Speed × Time

Key Points of the equation, Speed × Time = Distance.

1. When speed is constant, time is directly proportional to distance.
Time α Ddistance (when speed is constant.)

Illustration:
A car moves for 2 hours at a speed of 25 kmph and another car moves for 3 hours at the S₂me speed. Find the ratio of distances covered by two cars.
Since, the speed is constant, we can conclude that time α distance.
Hence, $$\frac{{{\text{Ta}}}}{{{\text{Tb}}}} = \frac{{{{\text{D}}_{\text{a}}}}}{{{{\text{D}}_{\text{b}}}}}$$
Since, the times traveled are 2 and 3 hours respectively; the ratio of distance covered is also $$\frac{2}{3}$$.

2. When time is constant speed is directly proportional to distance.
i.e. speed α distance. (When time is constant)
A body travels at S₁ kmph for the first 2 hours and then travels at S₂ kmph for the next two hours.
In such case the following ratios will be valid:
$$\frac{{{S_1}}}{{{S_2}}} = \frac{{{{\text{D}}_{\text{a}}}}}{{{{\text{D}}_{\text{b}}}}}$$

3. When distance is constant, speed is inversely proportional to time.
i.e. speed α $$\frac{1}{{{\text{time}}}}$$ .
In such case, the following ratios will be valid:
$$\frac{{{{\text{S}}_1}}}{{{{\text{S}}_2}}} = \frac{{{{\text{T}}_2}}}{{{{\text{T}}_1}}}$$

Illustrations:
A train meets with an accident and moves at $$\frac{3}{4}$$ of its original speed. Due to this, it is 20 minutes late. Find the original time for the journey beyond the point of accident.

Solution:
In the above case, distance of whole journey is constant.
Then, we can use, speed a 1/time.
Due to accident speed become $$\frac{3}{4}$$ then time becomes $$\frac{4}{3}$$.
Since, extra time = $$\frac{1}{3}$$ of normal time = 20 minutes.
Hence, normal time = 60 minutes.

Conversion between kmph to m/s:

1 km/h = 1000 m/s = $$\frac{{1000}}{{3600}}$$  m/s = $$\frac{5}{{18}}$$ m/s.
Hence, to convert y km/h into m/s multiply by $$\frac{5}{{18}}$$.
Thus, y km/h= $${\text{y}} \times \frac{5}{{18}}$$  m/s.
And vice-versa: y m/s = $${\text{y}} \times \frac{{{\text{18}}}}{5}$$  km/h. To convert from m/s to kmph, multiply by $$\frac{{{\text{18}}}}{5}$$.

Example 1:
A train is running with the speed of 45 km /h. What its speed in meter per second?
Solution:
Speed of train in meter/s = $$45 \times \frac{5}{{18}}$$  = 12.5 m/s.

Example 2:
If a motor car covers a distance of 250 m in 25 seconds, what is its speed in kilometer per hour?
Solution:
Speed of motor car = $$250 \times \frac{{18}}{5}$$  = 36 km/h.

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