PHYSICS PLZ HELP?
2020-08-21 12:48 pm
In an historical movie, two knights on horseback start from rest 82.8 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.373 m/s2, while Sir Alfred's has a magnitude of 0.220 m/s2. Relative to Sir George's starting point, where do the knights collide?
回答 (4)
2020-08-21 1:56 pm
Let x m be the distance of the place where they collide relative to Sir George' starting point.
s = ut + (1/2)at²
For u = 0: s = (1/2)at²
Hence, t² = 2s/a
For Sir George:
t² = 2x/0.373 …… [1]
For Sir Alfred:
t² = 2(82.8 - x)/0.220 …… [2]
[1] = [2]:
2x/3.73 = 2(82.8 - x)/0.220
x * 0.220 = (82.8 - x) * 0.373
0.220x = 30.8844 - 0.373x
0.593x = 30.8844
x = 52.1
They collide at a point 52.1 m relative to Sir George's starting point.
s = ut + (1/2)at²
For u = 0: s = (1/2)at²
Hence, t² = 2s/a
For Sir George:
t² = 2x/0.373 …… [1]
For Sir Alfred:
t² = 2(82.8 - x)/0.220 …… [2]
[1] = [2]:
2x/3.73 = 2(82.8 - x)/0.220
x * 0.220 = (82.8 - x) * 0.373
0.220x = 30.8844 - 0.373x
0.593x = 30.8844
x = 52.1
They collide at a point 52.1 m relative to Sir George's starting point.
2020-08-21 1:05 pm
Sir George's motion Sg = 0.373/2*t^2
Sir Alfred's motion Sa = 0.220/2*t^2
since Sg+Sa = 82.8 m, then :
t^2(0.373+0.220) = 82.8*2
t = √ 82.8*2/(0.373+0.220) = 16.7 sec
collision point dg = 0.373/2*16.7^2 = 52.1 m
Sir Alfred's motion Sa = 0.220/2*t^2
since Sg+Sa = 82.8 m, then :
t^2(0.373+0.220) = 82.8*2
t = √ 82.8*2/(0.373+0.220) = 16.7 sec
collision point dg = 0.373/2*16.7^2 = 52.1 m
2020-08-21 2:59 pm
the distances are proportional to the speed which is proportional to the acceleration.George has 0.373 units and Alfreds is 0.220
therefore George covers 0.373/(0.373+0.220) of the original distance = 0.373/(0.593) * 82.8 = 52.1 m
therefore George covers 0.373/(0.373+0.220) of the original distance = 0.373/(0.593) * 82.8 = 52.1 m
2020-08-21 2:03 pm
A is George place and B is Alfred place. Distance between them 82.8m. They meet at x from George. So we have
x= 1/2 0.373*t^2, (82.8 -x)= 1/2 0.220 t^2. The time is the same. divide one by the other. You have x/(82.8-x)= 0.373/0.220= 373/220. You have
220 x= (82.8-x)373. (220+373) x= 82.8*373, x=52.08 m from George.
x= 1/2 0.373*t^2, (82.8 -x)= 1/2 0.220 t^2. The time is the same. divide one by the other. You have x/(82.8-x)= 0.373/0.220= 373/220. You have
220 x= (82.8-x)373. (220+373) x= 82.8*373, x=52.08 m from George.
收錄日期: 2021-04-24 08:09:35
原文連結 [永久失效]:
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