# Linear Motion Questions and Answers - Form 2 Topical Mathematics

1. Two motorists Kinyua and Nyaboke travelled between two towns K and M which are 580km apart. Kinyua started from K at 6.20 a.m and traveled towards M at 90km/hr.Nyaboke started from town M 1 2/3 hours later and traveled towards town K at an average speed of 120km/h. At a small shopping centre along the way, Kinyua had a snack and car check for 20 minutes before proceeding
1. How far from town M did they meet?
2. At what time did they meet?
1. A rally driver starts from town M going to town k at 9.30a.m. If he averages 180km/hr, Calculate the distance from K and the time when the rally driver overtook Nyabok
2. The distance between two towns A and B is 150km. A car starts from town A at 10.00a.m and travels at an average speed of 80km/h towards B. A transit lorry travels from B at 10:15a.m towards town A at an average speed of 40km/h. At what time will the two vehicles meet?
3. The diagram below shows the speed-time graph for a bus traveling between two towns. The bus starts from rest and accelerates uniformly for 50seconds. It then travels at a constant speed for 150 seconds and finally decelerates uniformly for 100seconds.

Given that the distance between the two towns is 2700m, calculate the ;
1. maximum speed in km/h, the bus attained
2. acceleration
3. distance the bus traveled during the last 50seconds
4. time the bus takes to travel the first half of the journey
4. A cyclist covers a distance of 45 kilometres at a speed of 10km/h and a further 45 kilometres at 15km/h. Find his average speed for the journey
5. A lorry left town A for town B 1¼ hours before a car. The lorry and the car are traveling in the same direction at 80kmh-1 and 120kmh-1 respectively. After the overtake, the car moved for another 199/800 hours before reaching town B. Calculate:
1. The time the car took before overtaking the lorry completely
2. The distance between the two towns
3. The time the lorry will take to reach town B after the arrival of the car
6. A country bus left Nairobi at 10.45a.m and traveled towards Mombasa at an average speed of 60km/h. A matatu left Nairobi at 1:15p.m on the same day and traveled along the same road at an average speed of 100km/h. The distance between Nairobi and Mombasa is 500km.
1. Determine the time of the day when the matatu overtook the bus
2. Both vehicles continue towards Mombasa at their original speeds. How long had the matatu waited before the bus arrived?
7. Two passenger trains A and B which are 240m apart are travelling at 164km/h and 88km/h respectively approach on another one a straight railway line. Train A is 150m long and train B is 100m long. Determine the time in seconds that elapses before the two trains completely pass each other
8. A bus 5m long completely overtakes a trailer 15m long travelling in the same direction in 4.8. seconds. If the speed of the bus is 40 km/hr, determine the speed of the trailer in km/hr.
9. Find the LCM and GCD of the following numbers: 2 x 3 x 53 and 24 x 32 x 52
10. A boat sails from a point A to a point B upstream, a distance of 30 km and back to A in 3hrs 12 min. The current in the river is flowing at 5km/hr. Determine the speed of the boat in still water.
11. Two friends Ojwang and David live 40 km apart. One day Ojwang left his house at 9.00 a.m. and cycled towards David’s house at an average speed of 15 km/h. David left his house at 10.30 a.m. on the same day and cycled towards Ojwang’s house at an average speed of 25 km/h.
1. Determine ;
1. The distance from Ojwang’s house, where the two friends met.
2. The time they met.
3. How far Ojwang was from David’s house when they met.
2. The two friends took 10 minutes at the meeting point and they cycled to David’s house at an average speed of 12 km/h. Find the time they arrived at David’s house.
12. Mr. Kamau left town S at 6.00a.m and travels at an average speed of 24km/hr towards RMrs. Ronoh left town R to town S 10minutes later and arrived at 7.00a.m. If distance RS = 42km, find;
1. Where and when they will meet
2. The time Kamau arrived at R
3. If at 7.00a.m another traveler left S and travels towards R at speed twice that of Mrs. Ronoh, find where and when Mr. Kamau was overtaken by the traveler if so
13. A train 100m long travelling at 72km/hr, overtakes another train traveling in the same direction at 56km/hr and passes it completely in 54 seconds.
1. Find the length of the second train
2. Find also the time they would have taken to pass one another if they had been traveling at these speeds in opposite directions
14. An unskilled worker may either walk to work along a route 5km to take a bus journey of 7km. The average speed of the bus is 24km/hr faster than his average speed. Taking the average walking speed as x km/hr;
1. Write down expressions for time of the journey;
1. When walking
2. When using the bus
2. The journey by bus takes 36 minutes less than the journey on foot, find his walking speed
3. Hence find the time he takes to talk to work
15. At 1.50 p.m. a matatu is traveling at 80 km/h and it is 40 km behind a motorcycle traveling at 60 km/h.
1. After how long will the matatu overtake the motorcycle?
2. At what time will the matatu overtake the motorcycle?
16. A bus left Nairobi at 8:00a.m and traveled towards Kisumu at an average speed of 80km/h. At 8.30a.m, a car left Kisumu towards Nairobi at an average speed of 120km/hr. Given that the distance between Nairobi and Kisumu is 400km, Calculate:-
1. The time the car arrived in Nairobi
2. The time the two vehicles met
3. The distance from Nairobi to the meeting point
4. The distance of the bus from Kisumu when the car arrived in Nairobi
17. Two trucks A and B travelling at 28km/hr and 26km/hr respectively approach one another on straight road. Truck A is 10m long, while truck B is 15m long. Determine the time in seconds that elapses before the trucks completely pass each other

1. Distance covered by Kinyua in 12/3hrs
= 5 x 90 = 150km
Distance traveled by Nyaboke during the rest = (
1/3 x 120) = 40km
x = 390 – x
120x = 90(390 – x)
90     120
= 167.1km
Time = 167.1/90 = 1.86
8.33 + 1.86 = 10.19; they met at = 10.11a.m
580 – (150 + 167 .1) = 262.9km from M
Before the rally driver started, Nyaboke had traveled for 1½ hrs
(
3/2 x 120) = 180km
= x + 180
120       80
180x – 120x = 21600
x = 360km
Distance from K = 580 – (180 + 360)
x = 40km
Time = 540/180 = 3hrs
(9.30 + 3hrs) = 12.30p.m
2. Distance covered by the car after 15 min =( ¼ x 80)km = 20km
Distance covered together = 130km
Relative speed = (80 + 40) = 120km/h
Time taken to meet
= (130)/120 hrs
= 1hr 5 min
Time they met = 10:15 a.m + 1:05 = 11:20 a.m
3.
1. ½ X 50h + ½ X 100 h + 150h = 2700
225h = 2700
H = 2700/225= 12m/s
Maximum speed = 12 x 60 x 60/1000
= 43.2km/h
2. Acceleration = 12/50 m/s
=
6/25 m/s
3. ½ X 50 x 6
=150 m
4. Time for half of journey
½ X 12 (50 + t + t) = ½ X 2700
6(50 + 2t) = ½ X 2700
50 + 2t = 225
T = 225 - 50/2= 87.5
Total time
= 50 + 87.5 = 137.5 sec

4. Time taken at 10km
=45/10 = 4.5 hrs
Time taken at 15km/hr
45/15 = 3hrs
Total time taken = (4.5 + 3) = 7.5
(4.5 + 3) = 7.5 hrs
Average speed
=90/7.5
= 12km/hr
1. D = 5/4 x 80 + 50/1000
= 100.05km
Speed = 120 – 80 = 40km/h
T = D/S = 100.05/40
= 2.50125hours
2. D = S x T = 120 + 100.05/4000 + 199/800
= 120 x 11000/40000
= 330km
3. Total time = 330/80
= 41/8hrs
Time lapse = 41/8– 5/4 + 100.05/40000 + 199/800
= 41/8 – 4
= 1/8hrs
1. Distance traveled by bus before the matatu started off the journey is
Distance = speed x time
= 60 x 2 ½
= 150km
Relative speed = 100- 60 = 40km/hr
The matatu would cover the bus head start of 150km in 150/40 hrs = 3.75hrs = 3hrs 45 min
∴ The matatu will overtake the bus after 3hrs 45 minutes
This will be 1:15 + 3:45 = 5.00pm
2. Time taken by the matatu to complete the remaining 350km = 350/100 = 3 ½ hrs
= 3hours 30 minutes
Time taken by the bus to complete the remaining 350
=350/60 = 55/6 hrs = 5 hours 50 minutes
Matatu waits for 5hr 50min – 3hr 30 min = 2 hrs 20 min
5. Total distance = 100 + 140 + 150 = 490
Total speed = 88 + 164 = 252 km/hr
252 km/hr into m/h = 252 x 1000/3600 = 70m/h
Time taken = 490/70 = 7 sec
6. Distance = (5 + 15)m = 20m = 0.02km
S Bus = 40 km/h
Trailer = xkm/h
Relative speed = (40 – x) km/h
T = 4.8 sec. = 4.8h/3600
S = D/T
(40 – x) = 0.02/48/3600
0.02 x 3600/48
= 15 km/h
40 – x = 15
x = 25 km/h
7. L.C.M = 24 x 32 x 53 = 1800
GC.D. = 2 x 3 x 52 = 150
8. Total distance = 60 cm
Total time taken = 3 1/5 hrs
Let speed in still water be x km/h
Speed upstream = (x – 5) km/h
Speed downstream = (x + 5) km/h
30   +   30   = 16
x - 5   x + 5      5
30x – 150 + 30x + 150 = 16/5 (x2 – 25)
300x = 16x2 – 400
x= -5/4 or 20
∴Speed in still water is 20 km/hr
9. When David left, Ojwang had covered 15 x 3/2 = 22.5 km.
1. Remaining dist. = 40 – 22.5 = 17.5 km
Relative speed = 15 + 25 = 40 km/h
Time taken before meeting = 17.5/40 = 0.4375 hrs
Ojwang covered 15 x 0.437 = 5.5625 km
Distance from Ojwang’s house = 22.5 + 6.5625 = 29.0625 k
2. 0.4375 = 26 min 15 sec
∴
They met at 10.30 + 26.15
= 10.56. 15 am.
3. 40 – 29.0625 = 10.9375 km
1. Time taken = 10.9375/12 = 0.9115 hrs
= 54 min, 41 sec.
They arrived at 10.56. 15 + 54.41 + 10 min
= 12.00. 56 pm.
1. In 10minutes Kamau has travelled
10 x 24 = 6km
60
Distance left = 42 – 6 = 36km
Relating speed = 24 + 50.4k/hr
= 74.4km/hr
Time taken to meet = 42/74.4= 0.565hrs
= 34minutes
Time for meeting is 6.10 + 34 mins = 6.44a.m
34 x 50.4 = 28.56km from R or 13.44 from S
60
2. Kamau arrival time
42km = 1.75hrs
24km/hr
= 1hr .45 minutes
6.00a.m
+
1.45
7.45a.m
3. Mrs Ronoh speed = D/T
= 50.4km/hr
Twice = 50.4 x 2 = 100.8
7.00a.m, Mr. Kamau covered = 1x24= 24km
Retain speed = 100.8- 24 = 76.8km/hr
So
24 = 8.75
?   = 76.8
He was overtaken at 7.00 + 18.75 = 7.18am
At distance of D = S x t
= 100.8 x 189.75/60
31.5km from S or 10.5km from R
1. A gains on B at the rate of (72 – 56) Km/hr or 16km/h
in 1 hr A gains on B 16km
In 545 A gains on B
16 X 1000 X 54 m = 240
60 X 60
The sum of the lengths of the two trains is 240m but the length of the first train is 100m
The length of the second train is 140m
2. Relative speed = (72 + 56) km/h = 128km/hr
Distance between A and B decrease at the rate of 128km/hr
The distance decreases by 240m
60 X 60 X 240 s = 27 seconds
128 X 1000           4
= 6 ¾ s
10.
1.
1. Time = D/S
= 5/x hrs
2. Time = 7/x + 24 hrs
2. 5/x - 36/60 = 7/x+24
7        = 25 - 3x
x + 24 5x        5x
35x = 25x – 3x2 + 600 – 72x
3x2 + 82x – 600 = 0
(3x + 100) (x-6) = 0
x = -100/or 6
His speed = 6km/hr
3. Time = S x T
= 5/6 x 60
= 50 mins
1. Relative speed = 80 - 60
= 20 km/h
Time = 40/20 hrs
= 2 hrs
2. 1.50 p.m. = 13.50 hrs.
Time = 13.50 + 2 = 15.50 hrs
1. Nairobi   →    400km   →    Kisumu
Speed = 120km/h
Distance = 400km
Time taken = 400/120 = 10 = 3hrs 20min
8.30 + 3hrs 20min = 11:50a.m
2. at 8.30a.m distance covered by bus = ½ x 80 = 40km
Dist. Left = 360km speed = 200km/h
Time taken = 360/200 = 1hr 48mins
They met at 8:30+ 1hr 48mins
= 10:18a.m
3.  8 – 10.18a.m is 2hrs 18mins distance = 2 x 80 +18/60 x 80
=160 + 24km = 184 from Nairobi
4. car arrived in Nairobi after 3hrs 20mins
Bus traveled a time of 3hrs 20mins + 30mins
3hrs 50mins
Dist. = 3 x 80 + 50/60 x 80 = 240 + 662/3
Distance from Kisumu = 93 1/3 km
11. Total distance = 25m
Relative speed= 54km/hr
To m/s = 54 x 1000/60 x 60= 15/ms
Time they met = 25/15
= 12/3 sec

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