## Speed, Time and Distance

### Distance :-

Total length covered by any person or any object between two places is called distance.
or, Product of speed and time is called distance.

Distance = Speed × time

### Time :-

To cover a certain distance by a person or any object the duration in hours, minutes or seconds spent that duration is called the time.
Or, Distance per unit speed is called time.

 Time = Distance Speed

### Speed :-

When a distance is covered by a certain rate, that rate is called speed .
or, Distance per unit time is called speed.

 Speed = Distance Time

Relation between Speed (S), Time(T) and Distance (D) :-

 Speed = Distance Time

 Time = Distance Speed

Distance = Speed × Time

In the above relation, the unit used for measuring the distance(D) covered during the motion and the unit of time(T) duration to cover the distance(D) will be the same as in numerator and denominator respectively of the unit used for the speed.

 Speed = Distance Time

Ex- A car covers 477 km in 9 hours, then find the speed of the car.
Solution:- Given, Distance = 477 km
Time = 9 hours

 Speed = Distance Time

 Speed = 477 = 53 km/h 9

 Time = Distance Speed

Ex- A train covers a distance of 400 km with a speed of 50 km/h. What time is taken by the train to cover this distance?
Solution:- given, distance = 400 km
speed = 50 km/h

 Time = Distance Speed

 = 400 = 8 hours 50

Distance = Speed × Time

Ex- A car crosses a bridge with a speed of 85 km/h. What will be the length of the bridge, if the car takes 6 hours to cross the bridge?
Solution:- given, speed = 85 km/h
time = 6 hours
Distance = Speed × Time
= 85 × 6
= 510 km

### Basic Rule related to Speed, Time and Distance :-

Rule- Conversion of km/h to m/s and m/s to km/h

 1 km/h = 1 km = 1000 m 1 h 60 minutes

 = 1000 m = 10 60 × 60 sec 36

 ∴ 1 km/h = 5 m/s. 18

i.e. to convert km/h to m/sec, multiply by 5/18 and to convert m/sec to km/h multiply by 18/5.

Ex- Convert 90 km/h into m/s.

 Solution:- 90 km/h = 90 × 5 = 25 m/s 18

Ex- Convert 35 m/s to km/h.

 Solution:- 35 m / s = 35 × 18 = 126 km / h 5

Rule- If speed is kept constant, then the distance covered by an object is proportional to time.
Distance ∝ Time ( speed constant )

 D1 = D2 T1 T2

Ex- A person covers 60 km in 2 hours. What distance covered in 7 hours?
Solution:- Here, speed is kept constant.
then,

 ∴ D1 = D2 T1 T2

 ⇒ 60 = D2 2 7

⇒ D2 = 30 × 7
∴ D2 = 210 km

Rule- If time is kept constant, then the distance covered by an object is proportional to speed.
Distance ∝ Speed ( time constant )

 D1 = D2 S1 S2

Ex- A person covers a distance of 12 km, while at a speed of 6 km/h. How much distance he would cover in same time, if he walks at a speed of 9 km/h?
Solution:- Given, distance = 12 km
speed = 6 km/h
new speed = 9 km/h

 Time = distance = 12 = 2 hours speed 6

Distance = speed × time
= 9 × 2
= 18 km
By Rule
Here, D1 = 12, S1 = 6, S2 = 9

 ∴ D1 = D2 S1 S2

 ⇒ 12 = D2 6 9

 ⇒ 2 = D2 9

⇒ D2 = 2 × 9
∴ D2 = 18 km

Rule- If distance is kept constant, then the speed of a body is inversely proportional to time. i.e.,

 Speed ∝ 1 Time

S1T1 = S2T2 = S3T3 = ……

Ex- A person covers a certain distance with a speed of 12 km/h in 12 minutes. If he wants to cover the same distance in 4 minutes, What should be his speed?

 Solution:- Speed = Distance Time

 ⇒ 12 = D 12

 ⇒ 12 = D × 60 12

 ∴ D = 12 km 5

 New Speed = Distance new time

 = ( 12 / 5 ) ( 4 / 60 )

 = ( 12 × 60 ) = 12 × 3 = 36 km/h ( 5 × 4 )

By Rule
Here, S1 = 12 km/h, T1 = 12 minutes, T2 = 4 minutes
S1T1 = S2T2

 ⇒ 12 × 12 = S2 × 4 60 60

 ⇒ S2 = 12 × 12 × 60 60 4

∴ S2 = 36 km/h

Rule- When two bodies A and B are moving in opposite direction with speed x km/h and y km/h respectively, then the relative speed of two bodies is

Relative speed = ( x + y ) km/h

Ex- Two persons are moving in the directions opposite to each other. The speed of the both person are 7 km/h and 4 km/h, respectively. Find the relative speed of the two persons in respect of each other.
Solution:-
For opposite direction,
Relative speed = ( x + y ) km/h
= 7 + 4
= 11 km

Ex- Train takes 11 sec and 10 sec to cross two person who are walking in opposite direction of train at 4 km/h and 5 km/h respectively. Find speed of train.
Solution:- Let the speed of train be x km/h.
Relative speed = ( x + 4 ) km/h and ( x + 5 ) km/h

 Speed = D T

 or, x + 4 = D 11

or, 11 ( x + 4 ) = D
or, 11x + 44 = D ……(1)

 Speed = D T

 or, x + 5 = D 10

or, 10 ( x + 5 ) = D
or, 10x + 50 = D ……..(2)
From Eq.(1) and Eq.(2), we get
11x + 44 = 10x + 50
or, 11x - 10x = 50 - 44
or, x = 6 km/h
∴ speed of train = 6 km/h

Rule- When two bodies A and B are moving in same direction with speed x km/h and y km/h respectively, then the relative speed of two bodies is

Relative Speed = ( x - y ) km/h

Ex- Two trains are running in the same direction. The speeds of two trains are 40 km/h and 45 km/h, respectively. What will be the relative speed of second train with respect to first?
Solution:- For same direction,
Relative speed = ( x - y )
= 45 - 40
= 5 km/h

Ex- Train is moving with a speed of 72 km/h crosses a person moving in the same direction with a speed of 5 m/sec in 8 sec. Find the length of the train.
Solution:- Let the length of the train be x km.

 Speed of train = 72 km/h = 72 × 5 = 20 m/sec 18

Speed of person = 5 m/sec
Relative speed = 20 - 5 = 15 m/sec
Distance = Speed × time
= 15 × 8
= 120 m
∴ Length of train = 120 m

Rule- When a body travels with different speeds for different duration, then average speed of that body for the complete journey is defined as the ratio of total distance covered to the total time taken by body i.e.,

 Average Speed = Total distance traveled Total time taken

Ex- A person covers a distance of 35 km by bus in 40 min. After deboarding the bus, he took rest for 35 min and covers another 25 km by a taxi in 15 min. Find his average speed for the whole journey.
Solution:- Total distance = 35 + 25 = 60 km
Total time = 40 + 35 + 15 = 90 min = 90/60 = 3/2 hours

 Average Speed = Total distance traveled Total time taken

 = 60 = 60 × 2 3 / 2 3

= 40 km/h

Rule- If with two different speeds x and y the same distance is covered, then

 Time = distance speed

 T1 = D x

 T2 = D y

 Average Speed = Total distance traveled = ( D + D ) Total time taken ( T1 + T2 )

 = 2D { ( D / x ) + ( D / y ) }

 = 2D { D( x + y ) / xy }

 ∴ Average Speed = 2xy ( x + y )

Ex- A person goes from A to B with speed of 40 km/h and returns with a speed of 60 km/h. Find the average speed of the whole journey.
Solution:- Let the distance be D km.

 T1 = D 40

 T2 = D 60

 Average Speed = Total distance traveled Total time taken

 = ( D + D ) { ( D / 40 ) + ( D / 60 ) }

 = 2D { D( 60 + 40 ) / ( 40 × 60 ) }

 = ( 2 × 40 × 60 ) = 48 km/h 100

By Rule,
Let x = 40 and y = 60

 ∴ Average Speed = 2xy ( x + y )

 = ( 2 × 40 × 60 ) ( 40 + 60 )

 = ( 2 × 40 × 60 ) = 48 km/h 100

Rule- If with three different speeds x, y and z the same distance is covered, then average speed

 Time = distance speed

 T1 = D x

 T2 = D y

 T3 = D z

 Average Speed = Total distance traveled = ( D + D + D ) Total time taken ( T1 + T2 + T3 )

 = 3D { ( D / x ) + ( D / y ) + ( D / z ) }

 = 3D { D ( 1/x ) + ( 1/y ) + ( 1/z ) }

 ∴ Average Speed = 3xyz ( xy + yz + zx )

Ex- A covers 1/3rd of the journey at the speed of 10 km/h and half of the rest at the speed of 20 km/h and rest at the speed of 30 km/h. What is the average speed of A?
Solution:- Distance covered at 10 km/h = 1/3 of the whole journey

 Distance covered at 20 km/h = 1 - 1 × 1 3 2

 = 2 × 1 = 1 of the whole journey 3 2 3

 Distance covered at 30 km/h = 1 - 2 = 1 3 3

or the whole journey
Since the distances covered with each of the three given speeds are the same, therefore

 ∴ Average Speed = 3xyz ( xy + yz + zx )

 = ( 3 × 10 × 20 × 30 ) = ( 3 × 10 × 20 × 30 ) ( 10 × 20 + 20 × 30 + 30 × 10 ) ( 200 + 600 + 300 )

 = ( 3 × 10 × 20 × 30 ) = 180 km / h 1100 11

Rule- When a person covers a certain distance between two certain places with speed 'x', he gets his destination late by time t1 but when he covers the same distance with speed 'y', he reaches his destination t2 time earlier. In this case, the distance between two places is given by

 D = Product of speeds ( Difference of speeds × Difference of times )

 D = xy { ( x - y ) × ( t1 - t2 ) }

Ex- A covers a certain distance between his home and college by cycle. Having an average speed of 50 km/h, he is late by 25 min. However, with a speed of 40 km/h, he reaches his college 15 min earlier. Find the distance between his house and college.
Solution:- Let the distance between house and college be D km.

 Time = distance speed

 T + 25 = D 60 50

 or, T = D - 25 ......... ( 1 ) 50 60

 T - 15 = D 60 60

 or, T = D + 15 ......... ( 2 ) 60 60

From Eq.(1) and (2), we get

 D - 25 = D + 15 50 60 60 60

 or , D - D = 25 + 15 50 60 60 60

 or, 6D - 5D = 40 300 60

 or, D = 40 300 60

or, D = 40 × 5
∴ D = 200 km
∴ Distance between house and college = 200 km
By Rule,
x = 50 , y = 60, t1 = 25, t2 = 15

 D = xy { ( x - y ) × ( t1 - t2 ) }

 = ( 50 × 60 ) [ (60 - 50 ) × { ( 25 + 15 ) / 60 }]

 = ( 50 × 60 ) = 200 km { 10 × ( 40 / 60 ) }

Rule- If two cars started towards each other with a speed of x and y respectively. After crossing each other they took t1 and t2 time respectively to reach their destination, then ratio of their speed

Ratio of their speed = √y : √x

Ex- Two train started towards each other after passing each other they takes 16 hr and 9 hr respectively to reach their destination. Find the ratio of speeds of 1st train to that of 2nd train.
Solution:- Here, x = 16 hr, y = 9 hr
Ratio of 1st train to 2nd train = √y : √x
= √9 : √16
= 3 : 4

Rule- If excluding stoppage the speed of the train is x km/h and including stoppage speed of train is y km/h, then

 Stoppage time per hour = ( x - y ) x

Ex- Excluding stoppage the speed of bus is 54 km/h and including stoppage speed of bus is 45 km/h. Find for how many minutes does the bus stop per hour.
Solution:-
Here, x = 54 km/h and y = 45 km/h

 Stoppage time per hour = ( x - y ) x

 = ( 54 - 45 ) = 9 54 54

 = 1 hr = 1 hr = × 60 = 10 min/h 6 6

Rule- A policeman sees a thief at a distance of D. He starts chasing the thief who is running at a speed of 'x' and policeman is chasing with a speed of 'y'( y > x ). In this case, the distance covered by the thief when he is caught by the policeman, is given by

 Distance = D × x ( y - x )

Ex- A policeman sees a chain snatcher at a distance of 70 m. He starts chasing the chain snatcher who is running with a speed of 7 m/sec while the policeman chasing him with a speed of 9 m/s. Find the distance covered by the chain snatcher when he is caught by the policeman.
Solution:- Here, D = 70 m , x = 7 m/sec, y = 9 m/sec

 Distance = D × x ( y - x )

 = 70 × 7 ( 9 - 7 )

 = 70 × 7 2

= 35 × 7 = 245 m