【speedcrew】...spacewithaforwardspeedof53km/s.Tothecrew'sgreat....
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解决时间 2021-02-27 11:18
- 提问者网友：轮囘Li巡影
- 2021-02-26 10:49
【speedcrew】...spacewithaforwardspeedof53km/s.Tothecrew'sgreat....
最佳答案
- 五星知识达人网友：渊鱼
- 2021-02-26 12:11
【答案】 Enterprise's speed is V1=53km/s
Klingon's speed is V2=20km/s
let's suppose Enterprise's acceleration ：a=?km/s^2
suppose accelere time：t=?s
an Uniformly retarded rectilinear motion
→with Enterprise S1=V0t+1/2at^2=53*t+1/2at^2 ①
→with Klingon S2=Vt=20*t②
to barely avoid a collision
→S1=S2+140KM③
solve①,②and③
a≈-3.8893km/s^2
Anather way
if Enterprise can reduce the speed to 20km/s in the 140km,it can barely avoid a collision!
so this question became → S=1/2at^2
use average speed → V=[0+（53-20）]/2=16.5km/s
so S=V(average speed)*t → t=140/16.5
so acceleration a=△V/t=(-33)/(140/16.5)=-3.8893
Klingon's speed is V2=20km/s
let's suppose Enterprise's acceleration ：a=?km/s^2
suppose accelere time：t=?s
an Uniformly retarded rectilinear motion
→with Enterprise S1=V0t+1/2at^2=53*t+1/2at^2 ①
→with Klingon S2=Vt=20*t②
to barely avoid a collision
→S1=S2+140KM③
solve①,②and③
a≈-3.8893km/s^2
Anather way
if Enterprise can reduce the speed to 20km/s in the 140km,it can barely avoid a collision!
so this question became → S=1/2at^2
use average speed → V=[0+（53-20）]/2=16.5km/s
so S=V(average speed)*t → t=140/16.5
so acceleration a=△V/t=(-33)/(140/16.5)=-3.8893
全部回答
- 1楼网友：千杯敬自由
- 2021-02-26 13:19
这下我知道了
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