1) The average speed of A is 1.5 times that of B. B travels at 50 km/h. How long will it take A to catch up with B if B is 100 km ahead at first?
2) Two boats set off at 20 km/h southwards and 18 km/h towards north respectively .How far apart are they after 90 munutes?
3) A and B had a race. They ran at 7 m/s and 6 m/s respectively. How far apart were they after 2 minutes?
數學題幫忙列式
2007-07-27 8:45 pm
回答 (3)
2007-07-27 8:58 pm
✔ 最佳答案
1) Speed A catch up B = 0.5 x 50 km/h = 25km/hDistance apart = 100km
Time =100/25=4 hrs.
2) 90min=1.5hr
Speed of their leaving = 20+18=38km/h
Distance = 38 x 1.5 = 57km
3) Speed different = 7-6=1m/s
Distance = 1m/s x 2 min=1 x 2 x 60 = 120m
2007-07-27 9:06 pm
1)100divide(50x1.5)-50
=4hours
2)90divide20+90divide18
=9.5km
3)(60x2)divide(7-6)
=120m
hope can help u!
=4hours
2)90divide20+90divide18
=9.5km
3)(60x2)divide(7-6)
=120m
hope can help u!
參考: me
2007-07-27 9:04 pm
( 1 ) The speed of A : 50 km/h ( 1 . 5 ) = 75km / h
Let the required time be t h.
50t + 100 = 75t
25t = 100
t = 4
Hence 4 hours are required.
( 2 ) 90min = 1.5 h
So, distance travelled by boat A : ( 20 )( 1.5 ) = 30km
Distance travelled by boat B : ( 18 )( 1.5 ) = 27km
Then, by Pyth. Theorem, AB = 40.4km ( cor. to 3 s.f. )
( 3 ) 2min = 120s
Distance travelled by A : 7 ( 120 ) = 840m
Distance travelled by B : 6 ( 120 ) = 720m
Hence distance between them : 840 - 720 = 120m
Let the required time be t h.
50t + 100 = 75t
25t = 100
t = 4
Hence 4 hours are required.
( 2 ) 90min = 1.5 h
So, distance travelled by boat A : ( 20 )( 1.5 ) = 30km
Distance travelled by boat B : ( 18 )( 1.5 ) = 27km
Then, by Pyth. Theorem, AB = 40.4km ( cor. to 3 s.f. )
( 3 ) 2min = 120s
Distance travelled by A : 7 ( 120 ) = 840m
Distance travelled by B : 6 ( 120 ) = 720m
Hence distance between them : 840 - 720 = 120m
參考: My Maths Knowledge
收錄日期: 2021-04-23 17:11:13
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