سؤال

[زهرة البراري]

السلام عليكم

لدي سؤال واتمنى ان تسطيعوا الاجابة عليه

A spaceship ferrying workers to moon base i takes a straight-line oath from the earth to the moon, a distance of 384000 Km. Suppose it accelerates at 20 m/s2 for the first 15 min of the trip, then travels at constant speed until the last 15 min, when it accelerates at -20 m/s2 , just coming to rest as it reaches the moon.
a- what fraction of the total distance is traveled at constant speed
b- what total time is required for the trip

و اتمنى ان تردوا علي بأسرع وقت ممكن
و شكرا

[نوبل]

السلام عليكم ورحمة الله و بركاته.

D1 is the distance that traveled by the spaceship when it accelerates.
D2 is the Distance traveled at consistent speed.
D3 is the distance that traveled by the spaceship when it deaccelerates.
D = Total Distance= 384000 Km = 384*10^6.

D1 = D0 + V0*T  + ½*A*T^2  =  0+0+1/2*20*(15*60)^2 = 8.1*10^6 m
V = V0 + A*T = 0+20*(15*60) = 18*10^3 m/s
D3 = V1*T + ½*A*T^2 = (18*10^3)*(15*60) + ½*-20*(15*60)^2 = 16.2*10^6 – 8.1*10^6 = 8.1*10^6
D2 = D – (D1 +D3)  = 384*10^6 – (8.1*10^3 + 8.1*10^3) = 367.8*10^6 m.
D2/D = 367.8*10^6 / 384*10^6 = 0.958.
Constant speed Time = D2 / V = 367.8*10^6 / 18*10^3 = 20.43’*10^3 s
Total Time = Accelerate Time + Deaccelerate Time + Constant speed Time
T “total” = 15*60+ 20.34’*10^3 + 15*60 = 22.23’*10^3s = 6.18hour

[زهرة البراري]

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