As shown in a figure, there is a highway of the two lanes with same direction of movement.
A car runs at 80 km/h in A lane and at 110 km/h in B lane and when passing, it goes into B lane.
The car of ① ~ ⑩ is running A lane while maintaining the interval of 100 m (Fig. 1).
Now, the car ⑩ at the tail end began to run B lane (Fig. 2).
Next, when the car ⑩ runs with car ⑧ side by side, the car ⑨ began to run B lane (Fig. 3).
Similarly, when the car ⑩ runs with car ⑥ side by side, the car ⑧ begins to run B lane (Fig. 4).
Thus, while the car at the tail end of A lane goes into B lane one after another and maintains the interval of 100 m.
The car ⑩ goes back to A lane when it comes 100 m before the car ① (Fig. 5).
When the car ⑨, ⑧, and .... also come 100 m before the car of the head of A lane, it will go back to A lane.
A car runs at 80 km/h in A lane and at 110 km/h in B lane and when passing, it goes into B lane.
The car of ① ~ ⑩ is running A lane while maintaining the interval of 100 m (Fig. 1).
Now, the car ⑩ at the tail end began to run B lane (Fig. 2).
Next, when the car ⑩ runs with car ⑧ side by side, the car ⑨ began to run B lane (Fig. 3).
Similarly, when the car ⑩ runs with car ⑥ side by side, the car ⑧ begins to run B lane (Fig. 4).
Thus, while the car at the tail end of A lane goes into B lane one after another and maintains the interval of 100 m.
The car ⑩ goes back to A lane when it comes 100 m before the car ① (Fig. 5).
When the car ⑨, ⑧, and .... also come 100 m before the car of the head of A lane, it will go back to A lane.
The size of a car and the time of a lane change will not be taken into consideration.
Answer the following questions.
(1) Find the time while it will take from the situation of Fig. 2. to the situation of Fi.g 5.
(2) Find the time while the car ⑩ comes to the tail end and begins passing again from the situation of Fig. 2.
(3) Find the average speed per hour which the car ⑩ had run until it will be in the situation of (2) from the situation of Fig. 2.
Answer
(1) 2 minutes
(2) 4 minutes
(3) 95 km/h
Solution
(1) Fig. 5 is a figure where ⑩ at the tail end entered 100 m before ①.
It is that ⑩ moved 100 m × 10 = 1000m = 1 km.
It moves 1 km at 110 km/h - 80 km/h = 30 km/h.
It takes 1km / 30km/h = 1/30 hours. 60 minutes × 1/30 = 2 minutes.
(2) According to (1), it turned out it takes 2 minutes from Fig. 2 to Fig. 5.
Then, a figure when ⑨ runs 200 m and enters before ⑩ is Fig. 6.
In this case, ⑩ displaces laterally from the position of Fig. 5.
It moves 4 times (5 times in total) in the same way hereafter and as shown in Fig. 7, ⑩ comes to the tail end and will be in the situation of beginning to pass again.
As for the time, it takes 0.2 km / 30 km/h × 5 = 0.2/6 hour = 60 minutes × 0.2/6 = 2 minutes.
From the state of Fig. 2, it takes 2 minutes + 2 minutes = 4 minutes.
(3) Ask for the average speed of ⑩ for 4 minutes.
Average speed is calculated by the formula, (the total distance it ran) / (the time taken).
According to (2), ⑩ ran at 110 km/h for 2 minutes and at 80 km/h for 2 minutes.
110 km/h × 2/60 hours + 80 km/h × 2/60 hours = 193 km.
Therefore, the average speed to be found is 193km / 4/60 hour = 95km/h.
Answer the following questions.
(1) Find the time while it will take from the situation of Fig. 2. to the situation of Fi.g 5.
(2) Find the time while the car ⑩ comes to the tail end and begins passing again from the situation of Fig. 2.
(3) Find the average speed per hour which the car ⑩ had run until it will be in the situation of (2) from the situation of Fig. 2.
Answer
(1) 2 minutes
(2) 4 minutes
(3) 95 km/h
Solution
(1) Fig. 5 is a figure where ⑩ at the tail end entered 100 m before ①.
It is that ⑩ moved 100 m × 10 = 1000m = 1 km.
It moves 1 km at 110 km/h - 80 km/h = 30 km/h.
It takes 1km / 30km/h = 1/30 hours. 60 minutes × 1/30 = 2 minutes.
(2) According to (1), it turned out it takes 2 minutes from Fig. 2 to Fig. 5.
Then, a figure when ⑨ runs 200 m and enters before ⑩ is Fig. 6.
In this case, ⑩ displaces laterally from the position of Fig. 5.
It moves 4 times (5 times in total) in the same way hereafter and as shown in Fig. 7, ⑩ comes to the tail end and will be in the situation of beginning to pass again.
As for the time, it takes 0.2 km / 30 km/h × 5 = 0.2/6 hour = 60 minutes × 0.2/6 = 2 minutes.
From the state of Fig. 2, it takes 2 minutes + 2 minutes = 4 minutes.
(3) Ask for the average speed of ⑩ for 4 minutes.
Average speed is calculated by the formula, (the total distance it ran) / (the time taken).
According to (2), ⑩ ran at 110 km/h for 2 minutes and at 80 km/h for 2 minutes.
110 km/h × 2/60 hours + 80 km/h × 2/60 hours = 193 km.
Therefore, the average speed to be found is 193km / 4/60 hour = 95km/h.
It can be calculated more simply as (110 + 80) / 2 = 95 km/h in this case.
