Speed Time and Distance

 

Speed, Time and Distance

 

The basic concept of "speed, time and distance" are used in solving questions based on motion in straight line, relative, motion, problem from this chapter, basic concept should be, deeply understood because of the diverse range of problems.

Some important terms and formulae related to the chapter given below

 

SPEED

Speed is defined as distance covered by an object in a unit time interval. It is the rate at which distance is covered and

Obtained by dividing the distance covered by the object to the time it takes.

Unit of Speed km/h.

 

TIIME

Time is defined as quantity, which governs the order or sequence of an occurrence. In the absence of time, the actual sequence of any occurrence or incident would be lost. If we did not have the concept of time. We would not be able to know in what period or in what order something took place

Units of Time Hour (h), second (s), minutes (mm) etc.

 

DISTANCE

When an object is moving with a certain speed in a particular time the displacement made by an object is called the distance.

Units of Distance Miles, kilometre (km), metre (m) etc.

 

Relation between Speed, Time and Distance

                        \[\text{Speed=}\frac{\text{Distance}}{\text{Time}}\]

Note If the ratio of speeds of A and B is x : y, then the ratio of time taken by them to cover the same distance is given by

\[\frac{1}{x}:\frac{1}{y}\]

i.e.,                               \[y:x\]

 

Average Speed

It is the ratio of total distance covered to total time of journey.

 

Relative Speed

If two objects are in motion and their speeds are a and b, respectively. Then, their relative speed must be (a + b) when objects are moving in opposite direction.

Whereas their relative speed must be \[(a-b)\] when objects are moving in same direction.

 

Units Conversions

To convert km/h into m/s, we multiply the speed by \[\frac{5}{18}.\]

i.e.,                   \[1km/h=\frac{5}{18}m/s\]

To convert m/s into km/h, we multiply the speed by \[\frac{18}{5}.\]

i.e.,                   \[1\,m/s=\frac{18}{5}km/h\]

e.g.,      Convert 72 km/h into m/s.

Sol.      We know that, a km/h\[a\,km/h=\left( a\times \frac{5}{18} \right)m/s\]

\[\therefore \]      \[72\,km/h=\left( 72\times \frac{5}{18} \right)=4\times 5=20\,m/s\]

Quicker One

Ø   A person goes certain distance (A to B) at a speed of x km/h and returns back (B to A) at a speed of y km/h. The average speed during the whole journey is \[\frac{2xy}{x+y}km/h.\]

Ø   If a person convers \[{{d}_{1}}\] distance with speed \[{{s}_{1}},\] \[{{d}_{2}}\]distance with speed \[{{s}_{2}},\]\[{{d}_{3}}\]distance with speed \[{{s}_{3}}\] and so on. The average speed of the complete journey \[=\frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}+...}{\frac{{{d}_{1}}}{{{s}_{1}}}+\frac{{{d}_{2}}}{{{s}_{2}}}+\frac{{{d}_{3}}}{{{s}_{3}}}+...}\]

Average speed \[\text{=}\frac{\text{Total}\,\,\text{distance}\,\,\text{travelled}}{\text{Total}\,\,\text{time}\,\,\text{taken}}\]

Ø   When two persons A and B start at the same time from two points P and Q towards each other and after crossing each other they taken a h and b h in reaching Q and P, respectively.

Then,         \[\frac{\text{A }\!\!'\!\!\text{ s}\,\,\text{speed}}{\text{B }\!\!'\!\!\text{ s}\,\,\text{speed}}=\frac{\sqrt{b}}{\sqrt{a}}\]

Ø   If a person travels a certain distance by \[{{x}_{1}}\] km/h, then further the same distance by \[{{x}_{2}}\] km/h, then further the same distance by \[{{x}_{3}}\] km/h and so no, then the average speed for the whole journey is given by

Average speed \[=\frac{n}{\frac{1}{{{x}_{1}}}+\frac{1}{{{x}_{2}}}+\frac{1}{{{x}_{3}}}+...}km/h\]

Hence, n = Number of times the same distance travelled

Ø   If a person changes his speed to \[\frac{a}{b}\] of its usual speed and reaches early by T min, then the usual time taken by him is Usual time \[=\frac{T}{1-\frac{b}{a}}\left[ \frac{a}{b}>1 \right]\]

Ø   If two persons start walking from two different points with speed x km/h and y km/h respectively, then

(a) When they are walking in same direction, then after t h, they will by \[(x-y)\]t km apart

[Here, (x > y)]

(b) When they are walking in opposite direction, then after t h they will be \[(x+y)\] t km apart.

Ø   A thief is spotted by the policeman after t h. if the speed of thief A km/h and speed of policeman is B km/h (B > A), then time taken by the policemen to catch the thief is \[\left( \frac{tA}{B-A} \right)h.\]