KINEMATICS

Summary:

1. Kinematics deals with speed, distance and time

2. The following terms are computed as follows:

distance
Time =
(i) Distance = speed  time (ii) speed

distance total distance

Speed = Average speed =
(iii) time (iv) total time

3. Time in minutes can be converted into hours and vice versa as follows:
15
= = 0⋅25 hours
(i) 15 minutes 60

(ii) 025 hours = 025  60 = 15 minutes

−1 −1
4. Speed in kmh can be converted into m s as follows:

90 × 1000
90kmh−1= = 25m s−1
3600

−1
5. If a car travels at a constant speed of 60kmh , it means that the car covers
60km after every one hour

EXAMPLES:
−1
1. A car travels with a constant speed of 40kmh . How far can it travel in
21 minutes?

Soln:
21
= 40× = 14km
Distance = s  t 60

−1
2. A car travels 75km at a constant speed of 50kmh . How long does the journey
take?

1
Soln:
d 75
Time = = = 1⋅5h
s 50

3. A car travels 36km in 45minutes. Find its average speed

Soln:
d 45
Speed = = 36 ÷ = 48kmh− 1
t 60

−1
4. A car travels for 5 hours with a constant speed of 85kmh and then travels for
−1
3 hours with a constant speed of 69kmh . Find its average speed

Soln:
total distance
Average speed =
total time

Distance I = 85  5 = 425km

Distance II = 69  3 = 207km
425 + 207
Average speed= = 79kmh− 1
 5+ 3

−1
5. A car travels 975km with a constant speed of 65kmh and then travels 60km
−1
with a constant speed of 80kmh . Find its average speed

Soln:
total distance
Average speed =
total time

97⋅5 60
= = 1⋅5h = = 0⋅75h
Time I 65 Time II 80

2
 97⋅5 + 60
Average speed= = 70kmh− 1
 1⋅5 + 0⋅75

−1
6. A car travels for 5 hours at an average speed of 36kmh for the entire
−1
journey. For the first two hours its steady speed is 30kmh . Find its steady
speed for the last three hours

Soln:
total distance
Average speed =
total time

Distance I = 30  2 = 60km

Distance II = v  3 = 3v
60 + 3 v
36=
 5

−1
 v ¿ 40kmh
7. A man covers a distance of 15km in 3 hours, partly by walking and partly by

−1 −1
running. If he walks at 3kmh and runs at 9kmh , find the distance he

covers by running

Soln:

If x = distance walked, y = distance ran

⇒ x + y = 15 ------------( i )
x y
Also: + =3
3 9
⇒ 3x + y =27 -------------( ii )
On solving, y = 9km

3
8. Tom arrives early to school by 10 minutes when he rides from home at a steady
−1 −1
speed of 9kmh . When he rides at a steady speed of 7⋅5kmh , he arrives late

by 6 minutes. Calculate:

(i) how far the school is from his home

(ii) the speed that enables him to be punctual

Soln:

(i) let d = required distance, t = time required to be punctual

10
If difference in time = 60
d 1
⇒ t − = less time is used
9 6
∴ 54t − 6d = 9 ------------ ( i )
6
Also: If difference in time = 60
d 1
⇒ − t = more time is used
7⋅5 10
∴ 10d − 75t = 7⋅5 ------------( ii)

On solving, d = 12km, t = 1⋅5h

d 12
Speed = = = 8kmh−1
(ii) t 1⋅5

9. A train takes 15 minutes less for a journey of 156km if its speed is increased by
4kmh−1 from its normal speed. Find its normal speed

Soln:
15
let v = required speed If difference in time = 60

4
 156 156 1
⇒ − = more time is for slow speed
v v+ 4 4
2
∴ v + 4 v − 2496 = 0

v =
−4 ± √ 16 + 9984
2

v = 48 or −52

∴ v = 48kmh−1
C C
10. Two cyclists 1 and 2 left town P for town Q, 18km away at the same time.
C1 −1 C . C
travelled at a steady speed of 15kmh faster than 2 When 1 had covered
half the distance, he delayed for half an hour, after which he travelled at a speed
C .
20% less his original speed and arrived in town Q 15 minutes earlier than 2
C C
Determine the original speeds of the two cyclists 1 and 2

Soln:

C2
let v = speed of cyclist
18
Time to reach by C 2 =
v

9 1 9 2 v + 111
Time to reach by C 1 = + + =
v + 15 2 0⋅8( v + 15 ) 4( v + 15 )

15
If difference in time = 60

18 2 v + 111 1
⇒ − = more time is for slow speed
v 4( v + 15 ) 4
2
∴ v + 18 v − 360 = 0

v =
−18 ± √ 324 + 1440
2

v = 12 or −30

5
 ∴ v = 12kmh−1
C1 −1
 Speed of cyclist = v + 15 = 12 + 15 = 27kmh

EER:
−1
1. A car travels for 40 minutes with a constant speed of 84kmh . Find the speed
of another car which takes 48 minutes to travel the same distance
−1
 Ans: 70kmh 
−1
2. A car travels 97⋅5km with a constant speed of 65kmh and then travels for 45
−1
minutes with a constant speed of 80kmh . Find its average speed

−1
 Ans: 70kmh 
−1
3. Tom walking at 6kmh from home to school takes 20 minutes less when he
−1
returns at 10kmh . Calculate how far the school is from his home

 Ans: 5km 
−1 −1
4. Tom walks to work at 6kmh and returns home at 5kmh . If the entire
journey takes him 1 hour 39 minutes, calculate how far the place of work is from
his home

 Ans: 4⋅5km 
5. A car takes 15 minutes less for a journey of 70km if its speed is increased by
5kmh−1 from its normal speed. Find its normal speed
−1
 Ans: 35kmh 

6. A man covers a distance of 95km in 2 hours, partly by walking and partly by

6
 −1 −1
running. If he walks at 4kmh and runs at6kmh , find the distance he

covers by running

 Ans: 4⋅5km 

7. A train takes two hours less for a journey of 300km if its speed is increased by
5kmh−1 from its normal speed. Find its normal speed
−1
 Ans: 25kmh 
8. Towns P and Q are 156km apart. A car left P for Q at a steady speed of
Vkmh−1 . On the return journey, it increased the speed by 4kmh−1 and took
15 minutes less. Calculate the value of V
−1
 Ans: 48kmh 
9. It takes 3 hours to travel between two successive distances at respective speeds
−1 −1
of 50kmh and 60kmh . When the speeds are interchanged, the journey takes

8 minutes less. Calculate the distance of the entire journey

 Ans: 160km 
10. Kampala and Jinja are 300km apart. A car moves from Kampala to Jinja and
−1
back. Its average speed on the return journey is 30kmh greater than that on the
outward journey and it takes 50 minutes less. Find the average speed of the
outward journey
−1
 Ans: 90kmh 
11. Tom arrives early to school by 10 minutes when he rides from home at a steady
−1 −1
speed of 5⋅4kmh . When he rides at a steady speed of 3⋅6kmh , he arrives

late by 15 minutes. Calculate:

7
(i) how far the school is from his home

(ii) the speed that enables him to be punctual

−1
 Ans: (i) 4⋅5km (ii) 4⋅5kmh 

C C
12. Two cyclists 1 and 2 left town P for town Q, 24km away at the same time.
C1 −1 C . C
travelled at a steady speed of 10kmh faster than 2 When 1 had covered
half the distance, he delayed for three quarters of an hour, after which he travelled
at a speed 25% less his original speed and arrived in town Q 15 minutes earlier
C . C C
than 2 Determine the original speeds of the two cyclists 1 and 2
−1
 Ans: (a) 20kmh , 10kmh−1 (b)(i) 16km (ii) 08h

13. Towns P and Q are 130km apart. At 9:00am, a car left P for Q at a speed of
60kmh−1 and stopped at a petrol station for 10 minutes. It resumed its journey at
−1
a speed of 75kmh until it reached Q at 11:00am. Calculate:

(i) how far the petrol station is from town P

(ii) the average speed for the entire journey

−1
 Ans: (i) 30km (ii) 65kmh 
−1
14. A motorist travelled 8km up a hill at a steady speed of x kmh . On the
−1
return journey down the hill, his speed was ( x + 4 ) kmh . The difference in time

between the uphill and downhill journeys was 10 minutes.

(a) Write down an expression for the time taken for the:

(i) uphill journey

(ii) downhill journey

8
(b) (i) Form a quadratic equation for the difference in time for the two journeys

(ii) Solve the quadratic equation

(c) Find his average speed for the uphill and downhill journeys
2 −1 −1
 Ans: (b)(i) x + 4 x − 192 = 0 (ii) 12kmh (c) 13⋅7143kmh 

MOTION WITH CATCH UP OR MEETING

Summary:
In case of two bodies catching up or meeting during motion:
relative distance
Time to catch up =
(i) relative speed “Motion is in the same direction”
relative distance
Time to meet=
(ii) total speed “Motion is in opposite direction”

(iii) Relative speed is the same as speed difference

(iv) Relative distance is the same as distance apart at the start of timing

(v) Timing in this case starts with the latter rather than the former

EXAMPLES:
−1
1. A car is moving at 40kmh and a bus 30km behind it is moving in the same
−1
direction at 60kmh . Calculate the:

(i) time taken by the bus to catch up with the car

(ii) distance travelled by the bus to catch up with the car

Soln:

Bus 30km Car Catch up point

9
 relative distance 30
Time to catch up = = = 1⋅5h
(i) relative speed 60 − 40

(ii) Distance to catch up = s  t = 60  15 = 90km

Or Distance to catch up = 30 + (40  15) = 90km

METHOD 2

(i) Let t = time taken by the bus to catch up

Bus 30km Car 40t Catch up point

60t
Bus’s total distance = car’s total distance

60t = 30 + 40t

 t = 15h

(ii) Distance to catch up = s  t = 60  15 = 90km

Or Distance to catch up = 30 + (40  15) = 90km

−1
2. Tom left home riding steadily at 7kmh . Two hours later Bob left the same
−1
home riding steadily along the same road at 15kmh . Calculate:

(i) how long will it take Bob to catch up with Tom

(ii) how far will Bob travel to catch up with Tom

Soln:

Bob 14km Tom Catch up point

i) Tom’s distance in 2h = 7  2 = 14km
(

relative distance 14
Time to catch up = = = 1⋅75h
relative speed 15 − 7

(ii) Distance to catch up = s  t = 15  175 = 2625km

10
 Or Distance to catch up = 14 + (7  175) = 2625km

METHOD 2

(i) Let t = time taken by the Bob to catch up

Bob 14km Tom 7t Catch up point

15t

Bob’s total distance = Tom’s total distance

15t = (7  2) + 7t

 t = 175h

(ii) Distance to catch up = s  t = 15  175 = 2625km

Or Distance to catch up = 14 + (7  175) = 2625km

3. Bob and Tom have to go to church 125km away from their home. When Bob
−1
had covered 32km, riding steadily at 3kmh , Tom left the same home riding
−1
steadily along the same road at 5kmh .

(a) Calculate:

(i) how long will it take Tom to catch up with Bob

(ii) how far will Tom travel to catch up with Bob

(b) Immediately Tom caught up with Bob, he then reduced his speed and arrived
−1
035 hours later than if he had maintained the 5kmh speed.

(i) Calculate by how much he reduced his speed

(ii) For how long was he in church before Bob joined him

Soln:
125km
(a)
Tom 32km Bob Catch up Church
point
11
 relative distance 3⋅2
Time to catch up = = = 1⋅6h
(i) relative speed 5−3

(ii) Distance to catch up = s  t = 5  16 = 8km

(b) let v = new speed Remaining distance = 125  8 = 45km

If difference in time = 035

4⋅5 4⋅5
⇒ − = 0⋅35 more time is for slow speed
v 5

∴ v = 3⋅6kmh−1
−1
 Reduction in speed =5 − 3⋅6= 1⋅4kmh
4⋅5 − 4⋅5 =0⋅25h more time is for slow speed
(ii) Waiting time = 3 3⋅6

4. Towns P and Q are 500km apart. A car left P for Q at an average speed of
1
−1 2 hours ,
60kmh . After 2 a bus left P left for Q and travelled along the same
−1
road at an average speed of 100kmh .

(a) Calculate the:

(i) distance of the car from Q when the bus took off

(ii) distance from P to where the bus caught up with the car

(b) Immediately the bus caught up with the car, the bus stopped for 25 minutes.
Find the new average speed at which the bus travelled in order to reach Q at the
same time as the car

Soln:
500km
(a) P Q
Bus Car Catch up point

12
 1
2 h
(i) Car’s distance in 2 = 60  25 = 150km

 Distance from Q = 500  150 = 350km

relative distance 150
Time to catch up = = = 3⋅75h
(ii) relative speed 100 − 60

 Distance to catch up = s  t = 100  375 = 375km

(b) let v = new speed Remaining distance = 500  375 = 125km

If the total time to reach is the same

25 125 125
⇒ + =
60 v 60

∴ v = 75kmh−1
5. A car and a bus left town P for town Q 240km away at 8:00 am traveling at
90kmh−1 and 120kmh−1 respectively. After 20 minutes the bus stopped for
30 minutes and then resumed its journey at the same speed.

(a) Calculate the:

(i) time when the bus caught up with the car

(ii) distance from P to where the bus caught up with the car

(iii) time of arrival of the car to town Q

Soln:
500km
(a) P Q
Bus Car Catch up point

20
120 ×
(i) Bus’s distance in 20 minutes = 60 = 40km

13
 50
90 ×
Car’s distance in 50 minutes = 60 = 75km

relative distance 75 − 40 7
Time to catch up = = = h = 1h 10 minutes
relative speed 120 − 90 6

 Required time = 8:50 + 1h 10 minutes = 10:00 am

(ii) Distance from P = 40 + ( 120 × )

7
6 = 180km

240 8
= = h = 2h 40 minutes
(iii) Car’s travel time 90 3

 Arrival time = 8:00 + 2h 40 minutes = 10:40 am

6. Towns P and Q are 168km apart. A car left P for Q at an average speed of
60kmh−1 . At the same time a bus left Q for P at an average speed of 80kmh−1 .

(i) After how long will the two vehicles meet?

(ii) How far is the meeting point from town P?

−1
(iii) Just as they met, the car increased its speed by 16⋅8kmh . Find the
difference in the times of arrival of the two vehicles

Soln:
168km
P Q
Car Meeting point Bus

relative distance 168

Time to meet = = = 1⋅2h
(i) total speed 60 + 80

(ii) Distance from P = s  t = 60  12 = 72km

(iii) Bus’s distance to P = 72km

14
 72
= = 0⋅9h
 Bus’s time to P 80

−1
Car’s distance to Q = 168  72 = 96km New speed = 76⋅8kmh
96
= = 1⋅25h
 Car’s time to Q 76⋅8

 Arrival time difference = 125  09 = 035h

METHOD 2

(i) Let t = time taken by the Bus to meet

168km
P Q
Car Meeting point Bus
60t 80t
If the sum of their distances = 168

 60t + 80t = 168

 t = 12h

(ii) Distance from P = s  t = 60  12 = 72km

17. Bob and Tom live 62km apart. At 7:00 am, Bob left his home cycling towards
−1
Tom’s home at 20kmh . At 7:21 am, Tom left his home cycling towards Bob’s

−1
home at 24kmh .

(a) Calculate the:

(i) time when the two men met

(ii) distance from Bob’s house to where the two men met

(b) The two took 12 minutes at the meeting point and then travelled to Tom’s house
−1
at an average speed of 20kmh . Find the time they arrived at Tom’s house

Soln:

15
(a) (i)
7km 55km
P
Bob Meeting point Tom

21
20 ×
Bob’s distance in 21 minutes = 60 = 7km

relative distance 55
Time to meet = = = 1⋅25h = 1h 15minutes
 total speed 20 + 24

 Required time = 7:21 + 1h 15 minutes = 8:36 am

(ii) Distance from Bob’s house = 7 + (20  125 ) = 32km

(b) Distance to Tom’s house = 62  32 = 30km

30
= = 1⋅5h
 Their time to Tom’s house 20

 Arrival time = 8:36 + 12 + 1h 30 minutes = 10:18 am

METHOD 2

(i) Let t = time taken by Tom to meet

62km
P
Bob Meeting point Tom
7km 20t 24t
21
20 ×
Bob’s distance in 21 minutes = 60 = 7km

If the sum of their distances = 62

 ( 7 + 20t )+ 24t = 62

16
  t = 125h = 1h 15 minutes

 Required time = 7:21 + 1h 15 minutes = 8:36 am

(ii) Distance from Bob’s house = 7 + (20  125 ) = 32km

EER:

1. Bob and Tom have to go to school 9km away from their home. When Bob had
−1
covered 36km, walking steadily at 2⋅5kmh , Tom left the same home running
−1
steadily along the same road at 4kmh .

(a) Calculate:

(i) how long will it take Tom to catch up with Bob

(ii) how far will Tom travel to catch up with Bob

(iii) how long was Tom at school before Bob joined him

 Ans: (a)(i) 15h (ii) 6km (iii) 045h

2. Bob and Tom live 62km apart. At 7:00 am, Bob left his home cycling towards
−1
Tom’s home at 20kmh . At 8:00 am, Tom left his home cycling towards Bob’s

−1
home at 8kmh .

(a) Calculate the:

(i) time when the two men met

(ii) distance from Bob’s house to where the two men met

 Ans: (a)(i) 9:30am (ii) 50km 

17
3. Towns P and Q are 100km apart. At 4:00 am a cyclist left P for Q at a steady
−1
speed of 20kmh . At 7:30 am, a motorist left P for Q along the same road at a

−1
steady speed of 100kmh .

(a) Calculate the:

(i) time when the motorist overtook the cyclist

(ii) distance from P to where the motorist overtook the cyclist

(iii) time of arrival of the cyclist

 Ans: (a)(i) 8:22 am (ii) 875km (iii) 9:00 am 

4. Towns P and Q are 170km apart. At 8:25am car left P for Q at an average
−1
speed of 40kmh . At 8:55am bus left Q for P and travelled along the same road

−1
at an average speed of 80kmh .

(a) Calculate the:

(i) time when the two vehicles met

(ii) distance from P to where the two vehicles met

−1
(b) Just as they met, the car increased its speed by 10kmh . Find the difference
in their times of arrival at their destinations

 Ans: (a)(i) 10:10am (ii) 70km (b) 1125h

5. Bob and Tom have to go to church 308km away from their home. When Bob
−1
had covered 9km, riding steadily at 4kmh , Tom left the same home riding
−1
steadily along the same road at 7kmh .

(a) Calculate:

18
(i) how long will it take Tom to catch up with Bob

(ii) how far will Tom travel to catch up with Bob

(b) Immediately Tom caught up with Bob, he then reduced his speed and arrived
−1
06 hours later than if he had maintained the 7kmh speed.

(i) Calculate by how much he reduced his speed

(ii) For how long was he in church before Bob joined him
−1
 Ans: (a)(i) 3h (ii) 21km (b)(i) 2⋅1kmh (ii) 045h

6. Bob and Tom have to go for a burial 138km away from their home. When Bob
−1
had covered 18km, riding steadily at 24kmh , Tom left the same home riding
−1
steadily along the same road at 30kmh .

(a) Calculate:

(i) how long will it take Tom to catch up with Bob

(ii) how far will Tom travel to catch up with Bob

(iii) how long Tom will take waiting for Bob at the burial

(b) If Bob increased his speed immediately he was overtaken such that they both
arrive at the burial at the same time, calculate by how much he increased his speed
−1
 Ans: (a)(i) 3h (ii) 90km (b)(i) 2⋅1kmh (ii) 045h

7. Bob and Tom live 190km apart. At 7:00 am, Bob left his home cycling towards
−1
Tom’s home at 30kmh . At 7:30 am, Tom left his home cycling towards Bob’s

−1
home at 40kmh .

(a) Calculate the:

19
(i) time when the two men met

(ii) distance from Bob’s house to where the two men met

(b) The two took 15 minutes at the meeting point and then travelled to Tom’s house
−1
at an average speed of 20kmh . Find the time they arrived at Tom’s house

 Ans: (a)(i) 10:00am (ii) 90km (b) 3:15pm 

C C
8. Two cyclists 1 and 2 left town P for town Q, 24km away at the same time.
C1 −1 C . C
travelled at a steady speed of 10kmh faster than 2 When 1 had covered
half the distance, he delayed for three quarters of an hour, after which he travelled
at a speed 25% less his original speed and arrived in town Q 15 minutes earlier
C .
than 2

C1 C2
(a) Determine the original speeds of the two cyclists and

C C
(b) If cyclist 1 started from town P while 2 at the same time started from town
Q and both travelled nonstop,

(i) find how far from P the two cyclists will meet

(ii) After how long will they meet?

−1
 Ans: (a) 20kmh , 10kmh−1 (b)(i) 16km (ii) 08h

DISTANCETIME GRAPHS

EXAMPLES:

20
1. Towns P and Q are 500km apart. At 8:15 am a car left P for Q traveling at a
−1
steady speed of 60kmh . Two and a half hours later, a bus left P for Q along the

−1
same road at a steady speed of 100kmh .

(a) On the same axes show the journeys of the two vehicles
Use a scale of 2cm to represent 50km and 2cm to represent 1 hour 
(b) Use your graphs to find the:

(i) distance of the car from Q when the bus took off

(ii) time and distance from P where the bus overtook the car

(iii) difference in the times of arrival of the two vehicles

Soln:

Table for the car (P to Q)

Distance moved 0 60 120

Time of the day 8:15 9:15 10:15
Table for the bus (P to Q)

Distance moved 0 100 200

Time of the day 10:45 11:45 12:45

2. Towns P and Q are 360km apart. At 7:30 am a car left P for Q traveling at a
−1
steady speed of 80kmh . At the same time a bus left Q for P at an average speed

−1
of 100kmh .

(a) On the same axes show the journeys of the two vehicles

Use a scale of 2cm to represent 50km and 2cm to represent 1 hour 

(b) Use your graphs to find the:

21
 (i) time when the two vehicles met

(ii) distance from Q to where the two vehicles met

(iii) difference in the times of arrival of the two vehicles

Soln:

Table for the car (P to Q)

Distance moved 0 80 160

Time of the day 7:30 8:30 9:30
Table for the bus (Q to P)

Distance moved 0 100 200

Time of the day 7:30 8:30 9:30

3. Towns P and Q are 180km apart. At 0730 hours a car left P for Q traveling at a
1
−1 1 hours
steady speed of 40kmh . After 2 hours the car stopped for 2 and then
−1
proceeded with its journey at a speed of 50kmh . A bus left Q for P at the same
−1
time as the car at a steady speed of 60kmh but suddenly reduced its speed after
−1
2 hours to 15kmh for the rest of its journey

(a) On the same axes show the journeys of the two vehicles

Use a scale of 2cm to represent 20km and 2cm to represent 1 hour 

(b) Use your graphs to find the:

(i) time and distance from Q where the two vehicles met

(ii) distance between the two vehicles at 0930 hours

(iii) times of arrival of the two vehicles at their destinations

(iv) difference in the times of arrival at the respective towns

22
Soln:

Table for the car (P to Q)

Distance moved 0 40 80 80 130 180

Time of the day 0730 0830 0930 1100 1200 1300
Table for the bus (Q to P)

Distance moved 0 60 120 135 150 165

Time of the day 0730 0830 0930 1030 1130 1230

4. Towns P and Q are 150km apart. At 1100 hours a car left P for Q traveling at a
−1
steady speed of 50kmh . After half an hour a bus left P for Q at a steady speed

−1
of 120kmh but after traveling 30km, it stopped for 15 minutes and then
−1
resumed its journey at a speed of 120kmh . The bus arrived at Q and rested for
30 minutes before returning to P by the same road where it arrived at 1442 hours

(a) On the same axes show the journeys of the two vehicles

Use a scale of 2cm to represent 25km and 4cm to represent 1 hour 

(b) Use your graphs to find the:

(i) time and distance from Q where the bus overtook the car on its way to Q

(ii) time and distance from Q where the bus met the car on its way back to P

(iii) average speed of the bus for the outward journey

(iv) average speed of the bus for the return journey

(v) average speed of the bus for the entire journey

Soln:

23
 Table for the car (P to Q)

Distance moved 0 50 100

Time of the day 1100 1200 1300
Table for the bus (P to Q and back)

Distance moved 0 30 30 150 150 0

Time of the day 1130 1145 1200 1300 1330 1442
30
Time to cover 30km = = 0⋅25h = 15 minutes
120

EER:

1. Towns P and Q are 360km apart. At 8:15 am a car left P for Q traveling at a
2
−1 1 hours,
steady speed of 90kmh . After 5 a bus left P for Q along the same road
−1
at a steady speed of 120kmh .

(a) On the same axes show the journeys of the two vehicles
Use a scale of 2cm to represent 50km and 2cm to represent 1 hour 
(b) Use your graphs to find the:

(i) distance of the car from Q when the bus took off

(ii) time and distance from P to where the bus overtook the car

(iii) difference in the times of arrival of the two vehicles

2. Towns P and Q are 100km apart. At 5:00am, a car left P and travelled for one
−1 −1
hour at a speed of 30kmh . It then increased its speed to 100kmh until it
reached Q. At 5:30am, a bus left Q for P and travelled at a steady speed of
1
−1 1 hours
60kmh until it broke down 2 later.

24
(a) On the same axes show the journeys of the two vehicles

Use a scale of 2cm to represent 10km and 2cm to represent 30 minutes 

(b) Use your graphs to find the:

(i) time and distance from P when the two vehicles met

(ii) distance from Q to where the bus broke down

(iii) time the car reached town Q

3. Towns P and Q are 200km apart. At noon, a car left P and travelled for one
−1
hour at a speed of 50kmh . It stopped for 30 minutes then continued to Q at a

−1
speed of 60kmh . At 12:30pm, a bus left Q and travelled for one hour at a speed

−1 −1
of 40kmh . It then changed and travelled at a speed of V kmh and arrived at
4:30pm at town P

(a) On the same axes show the journeys of the two vehicles

Use a scale of 2cm to represent 20km and 4cm to represent 1 hour 

(b) Use your graphs to find the:

(i) time when the two vehicles met

(ii) distance from P to where the two vehicles met

(iii) time of arrival of the car

(iv) speed V of the bus

4. Towns P and Q are 360km apart. At 7:00am, a car left P and travelled for two
−1
hours at a speed of 50kmh . It stopped for 1 hour then continued to Q at a

25
steady speed for 4 hours. At 8:00am, a bus left Q for P and travelled nonstop for
1
4 hours .
2

(a) On the same axes show the journeys of the two vehicles

Use a scale of 2cm to represent 40km and 2cm to represent 1 hour 

(b) Use your graphs to find the:

(i) time when the two vehicles met

(ii) distance from P when the two vehicles met

(iii) average speed of the bus

5. Towns P and Q are 45km apart. At 0815 hours, Bob left P for Q riding at a
−1
speed of 15kmh . His bicycle broke down at 0915 hours and was delayed for 45

minutes. He then walked back to P and arrived at 1230 hours. At 0915 hours Tom
left P for Q riding at a steady speed and arrived at 1200 hours.

(a) On the same axes show the journeys of the two men

Use a scale of 2cm to represent 20km and 4cm to represent 1 hour 

(b) Use your graphs to find the:

(i) distance from P when Bob’s bicycle broke down

(ii) speed at which Bob walked back to P

(iii) average speed of Tom

(iv) time when the two men met

(iv) distance from P when the two men met

26
6. Towns P and Q are 90km apart. At 7:00 am, a car left P for Q traveling at a
−1
steady speed of 24kmh . 45 minutes later, a bus left P for Q at a steady speed of

60kmh−1 but after traveling 15km, it stopped for half an hour and then resumed
−1
its journey at a speed of 60kmh . The bus arrived at Q and rested for 15
minutes before returning to P by the same road where it arrived at 11:15 am.

(a) On the same axes show the journeys of the two vehicles

Use a scale of 2cm to represent 10km and 4cm to represent 1 hour 

(b) Use your graphs to find the:

(i) time and distance from P where the bus overtook the car on its way to Q

(ii) time and distance from Q where the bus met the car on its way back to P

(iii) average speed of the bus for the return journey

7. Towns P and Q are 450km apart. At 7:42am, a Van and a Bus left P for Q
−1 −1
travelling at 90kmh and 150kmh respectively. After 30 minutes, the bus
had a puncture which took 1×8 hours to mend before resuming the journey at
the same speed.
(a) On the same axes show the journeys of the two vehicles

Use a scale of 2cm to represent 50km and 2cm to represent 1 hour 

(b) Use your graph, to find the:

(i) distance from P to where the two vehicles met for the first time.

(ii) time and distance from Q to where the two vehicles met for the second time

(iii) difference in the times of arrival of the two vehicles

(iv) average speed of the bus for the entire journey

27
8. Town P is 300km from town Q. A lorry left town P for Q at 7:30am and
−1
travelled at a steady speed of 80kmh . At the same time, a bus left town Q
−1
for town P and travelled at a steady speed of 120kmh .
(a) On the same axes show the journeys of the two vehicles

Use a scale of 2cm to represent 50km and 2cm to represent 1 hour 

(b) Use your graph, to find the time and distance from Q to where the two vehicles
met
(c) Just as they met, the lorry and the bus were then driven at speeds of
100kmh−1 and 75kmh−1 respectively, calculate the difference in their times of
arrival at their destinations.

28