1. If you travel 7000 m at a rate of 25 m/s and, how long does it take to complete the trip? 2. How long does it take to get back?
回答 (6)
2020-09-02 12:00 pm
1.
Time taken for the trip
= Distance / Rate
= 7000/25 s
= 280 s
= 4 min 40 s
2.
If the rate is unchanged, time taken to get back
= Distance / Rate
= 7000/25 s
= 280 s
= 4 min 40 s
3.
Total time for the trip there and back
= 280 × 2 s
= 560 s
= 9 min 20 s
Time taken for the trip
= Distance / Rate
= 7000/25 s
= 280 s
= 4 min 40 s
2.
If the rate is unchanged, time taken to get back
= Distance / Rate
= 7000/25 s
= 280 s
= 4 min 40 s
3.
Total time for the trip there and back
= 280 × 2 s
= 560 s
= 9 min 20 s
2020-09-03 12:13 am
7000m is your 'given', start there:
7000m (1sec/25m) = 280 s
There and back just doubles the time: 560 s
((note: in KMS system, time is reported in seconds))
7000m (1sec/25m) = 280 s
There and back just doubles the time: 560 s
((note: in KMS system, time is reported in seconds))
2020-09-02 4:43 pm
time for getting there = tf = 7000 m / 25 m/sec = 280 sec
time for getting back = tr = 7000 m / 25 m/sec = 280 sec
total trip time t = tf+tr = 280+280 = 560 sec (a bit less than 10 min)
time for getting back = tr = 7000 m / 25 m/sec = 280 sec
total trip time t = tf+tr = 280+280 = 560 sec (a bit less than 10 min)
2020-09-02 1:39 pm
I must make a comment on the answer given. How are you going to go from +25 m/s to -25 m/s without taking some time to do so? There will be a discrepancy between what is possible and what the question asks.
2020-09-17 1:08 am
7000/25=280s.
If the speed is the same 280s.
total time 560s= 9.33min.=9m 19.8s.
If the speed is the same 280s.
total time 560s= 9.33min.=9m 19.8s.
2020-09-03 3:23 am
Do you know, I was not good at maths at school in the 1960s, but I could do that sum in my head, and write down my workings just to make sure.
收錄日期: 2021-04-24 07:58:46
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