There is an annular road as shown in the figure passing along P point and Q point.
The distance between P and Q is the same in a clockwise rotation or a counterclockwise rotation.
Taro moves from P clockwise and Jiro moves from Q counterclockwise on the road.
They leave simultaneously and continue moving by a fixed speed respectively.
5 minutes after leaving, Taro and Jiro met for the first time at R point since they left.
Taro moved 4 km after the 1st meeting and met with Jiro again at S point.
The distance of S point and Q point is 1.2 km along the road.
Answer each of following questions.
(1) Find the speed of Taro an hour.
(2) Find the distance of the shorter one along the road from P point of R point.
(3) Three kinds are considered as the distance between P and Q along the road and the speed of Jiro an hour.
Find both of them in all cases.
The distance between P and Q is the same in a clockwise rotation or a counterclockwise rotation.
Taro moves from P clockwise and Jiro moves from Q counterclockwise on the road.
They leave simultaneously and continue moving by a fixed speed respectively.
5 minutes after leaving, Taro and Jiro met for the first time at R point since they left.
Taro moved 4 km after the 1st meeting and met with Jiro again at S point.
The distance of S point and Q point is 1.2 km along the road.
Answer each of following questions.
(1) Find the speed of Taro an hour.
(2) Find the distance of the shorter one along the road from P point of R point.
(3) Three kinds are considered as the distance between P and Q along the road and the speed of Jiro an hour.
Find both of them in all cases.
Answer
Moreover, the sum of distance which two persons moved by the time they met again is the distance for one round of the road which is twice of the distance for 1st meeting..
That is, since the sum of the distance they moved was twice, the time taken by the 2nd meeting is also twice.
It turns out that it is 5 minute × 2 = 10 minutes.
Since it means that Taro moved 4 km from R point to S point in 10 minutes, the speed of Taro is 4 km / (10/60) = 24 km/h.
(3) As for the positional relationship among four points P, Q, R, S, there are two cases as shown in Fig.2 and Fig 3 in addition to the case of Fig.1.
① Case of Fig.1The distance between Q and R is 4 km - 1.2 km = 2.8 km.
Since Jiro moved this distance in 5 minutes, the speed of Jiro is 2.8 km / (5/60) = 33.6 km/h.
The distance between P and Q is 2 + 2.8 = 4.8 km.
② Case of Fig.2
The distance between Q and R = RS + SQ = 4 + 1.2 = 5.2 km.
Since Jiro moved this distance in 5 minutes, the speed of Jiro is 5.2km / (5/60) = 62.4 km/h.
Since PR = 2 km, the distance between P and Q is 2 + 5.2 = 7.2 km.
③ Case of Fig.3
QS = 1.2 km Since Jiro moved this distance in 5 +10 = 15 minutes, the speed of Jiro is 1.2km / (15/60) = 4.8 km/h.
PR = 2 km QR is the distance Jiro moved in 5 minutes.
QR = 4.8 km/h × 5/60 = 0.4 km.
The distance between P and Q is 2 + 0.4 = 2.4 km.
(1) 24 km/h
(2) 2 km
(3) 4.8 km and 33.6 km/h,
(2) 2 km
(3) 4.8 km and 33.6 km/h,
7.2 km and 62.4 km/h,
2.4 km and 4.8 km/h
Solution
(1) Since they met for the first time 5 minutes after leaving as shown in Fig. 1, the sum of distance which they moved in 5 minutes is equal to the distance between P and Q.
Moreover, the sum of distance which two persons moved by the time they met again is the distance for one round of the road which is twice of the distance for 1st meeting..
That is, since the sum of the distance they moved was twice, the time taken by the 2nd meeting is also twice.
It turns out that it is 5 minute × 2 = 10 minutes.
Since it means that Taro moved 4 km from R point to S point in 10 minutes, the speed of Taro is 4 km / (10/60) = 24 km/h.
(2) The distance between P point and R point is that Taro moved in five minutes which is 24 km/h × 5/60 hours = 2 km.
(3) As for the positional relationship among four points P, Q, R, S, there are two cases as shown in Fig.2 and Fig 3 in addition to the case of Fig.1.
① Case of Fig.1The distance between Q and R is 4 km - 1.2 km = 2.8 km.
Since Jiro moved this distance in 5 minutes, the speed of Jiro is 2.8 km / (5/60) = 33.6 km/h.
The distance between P and Q is 2 + 2.8 = 4.8 km.
② Case of Fig.2
The distance between Q and R = RS + SQ = 4 + 1.2 = 5.2 km.
Since Jiro moved this distance in 5 minutes, the speed of Jiro is 5.2km / (5/60) = 62.4 km/h.
Since PR = 2 km, the distance between P and Q is 2 + 5.2 = 7.2 km.
③ Case of Fig.3
QS = 1.2 km Since Jiro moved this distance in 5 +10 = 15 minutes, the speed of Jiro is 1.2km / (15/60) = 4.8 km/h.
PR = 2 km QR is the distance Jiro moved in 5 minutes.
QR = 4.8 km/h × 5/60 = 0.4 km.
The distance between P and Q is 2 + 0.4 = 2.4 km.
