- 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
- How far from town M did they meet?
- At what time did they meet?
- 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
- 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?
- 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 ;- maximum speed in km/h, the bus attained
- acceleration
- distance the bus traveled during the last 50seconds
- time the bus takes to travel the first half of the journey
- 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
- 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:
- The time the car took before overtaking the lorry completely
- The distance between the two towns
- The time the lorry will take to reach town B after the arrival of the car
- 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.
- Determine the time of the day when the matatu overtook the bus
- Both vehicles continue towards Mombasa at their original speeds. How long had the matatu waited before the bus arrived?
- 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
- 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.
- Find the LCM and GCD of the following numbers: 2 x 3 x 53 and 24 x 32 x 52
- 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.
- 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.
- Determine ;
- The distance from Ojwang’s house, where the two friends met.
- The time they met.
- How far Ojwang was from David’s house when they met.
- 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.
- Determine ;
- Mr. Kamau left town S at 6.00a.m and travels at an average speed of 24km/hr towards R. Mrs. Ronoh left town R to town S 10minutes later and arrived at 7.00a.m. If distance RS = 42km, find;
- Where and when they will meet
- The time Kamau arrived at R
- 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
- 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.
- Find the length of the second train
- Find also the time they would have taken to pass one another if they had been traveling at these speeds in opposite directions
- 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;
- Write down expressions for time of the journey;
- When walking
- When using the bus
- The journey by bus takes 36 minutes less than the journey on foot, find his walking speed
- Hence find the time he takes to talk to work
- Write down expressions for time of the journey;
- 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.
- After how long will the matatu overtake the motorcycle?
- At what time will the matatu overtake the motorcycle?
- 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:-
- The time the car arrived in Nairobi
- The time the two vehicles met
- The distance from Nairobi to the meeting point
- The distance of the bus from Kisumu when the car arrived in Nairobi
- 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
Answer
- 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 = 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 - 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 -
- 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 - Acceleration = 12/50 m/s
=6/25 m/s - ½ X 50 x 6
=150 m - 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
- ½ X 50h + ½ X 100 h + 150h = 2700
- 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 - D = 5/4 x 80 + 50/1000
= 100.05km
Speed = 120 – 80 = 40km/h
T = D/S = 100.05/40
= 2.50125hours - D = S x T = 120 + 100.05/4000 + 199/800
= 120 x 11000/40000
= 330km - Total time = 330/80
= 41/8hrs
Time lapse = 41/8– 5/4 + 100.05/40000 + 199/800
= 41/8 – 4
= 1/8hrs
- D = 5/4 x 80 + 50/1000
- 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 - 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
- Distance traveled by bus before the matatu started off the journey is
- 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 - 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 - L.C.M = 24 x 32 x 53 = 1800
GC.D. = 2 x 3 x 52 = 150 - 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 - When David left, Ojwang had covered 15 x 3/2 = 22.5 km.
- 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 - 0.4375 = 26 min 15 sec
∴ They met at 10.30 + 26.15
= 10.56. 15 am. - 40 – 29.0625 = 10.9375 km
- Remaining dist. = 40 – 22.5 = 17.5 km
- 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.
- 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 - Kamau arrival time
42km = 1.75hrs
24km/hr
= 1hr .45 minutes
6.00a.m
+
1.45
7.45a.m - 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
- In 10minutes Kamau has travelled
- 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 - 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
- A gains on B at the rate of (72 – 56) Km/hr or 16km/h
-
-
- Time = D/S
= 5/x hrs - Time = 7/x + 24 hrs
- Time = D/S
- 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 = -100/3 or 6
His speed = 6km/hr - Time = S x T
= 5/6 x 60
= 50 mins
-
- Relative speed = 80 - 60
= 20 km/h
Time = 40/20 hrs
= 2 hrs - 1.50 p.m. = 13.50 hrs.
Time = 13.50 + 2 = 15.50 hrs
- Relative speed = 80 - 60
- Nairobi → 400km → Kisumu
Speed = 120km/h
Distance = 400km
Time taken = 400/120 = 10 = 3hrs 20min
8.30 + 3hrs 20min = 11:50a.m - 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 - 8 – 10.18a.m is 2hrs 18mins distance = 2 x 80 +18/60 x 80
=160 + 24km = 184 from Nairobi - 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
- Nairobi → 400km → Kisumu
- 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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