物体在斜面顶端由静止开始匀加速下滑,在最初2秒内的位移为S1，在最后2秒内的位移为S2. 高一物理！！
\u4e00\u7269\u4f53\u4ece\u659c\u9762\u7531\u9759\u6b62\u5f00\u59cb\u5300\u52a0\u901f\u4e0b\u6ed1\uff0c\u6700\u521d3\u79d2\u5185\u7684\u4f4d\u79fb\u662fS1\uff0c\u6700\u540e3\u79d2\u4f4d\u79fb\u662fS2\uff0c\u82e5S1\uff1aS2=3\uff1a7\uff0cS2-S1=6\u7531S1\uff1aS2=3:7 S2-S1=6 \u5f97S1=9/2 S2=21/2 \u53c8S1=1/2 at*t a=1
\u8bbe\u6700\u540e3\u79d2\u524d\u901f\u5ea6\u4e3av vt+1/2 at*t=s2 v=2 \u524d\u9762\u5171\u6ed1\u884c\u4e86V=at t=2 \u603b\u5171\u6ed1\u884c\u4e862+3=5\u79d2\u52191/2 a t*t=25/2
\u4ec1\u5144\u786e\u5b9a\u6ca1\u9519\uff0c\u6c42\u51faa=1\uff0cS2\u59cb\u4f4d\u7f6e\u901f\u5ea6\u4e3a2\uff0c\u8fd9\u600e\u4e48\u53ef\u80fd\u5462\uff1f\uff1f\uff1f
设：加速度为：a，总时间为：t，斜面长度为：L则有：s1=at1^2/2，t1=2s，可得：s1=2a，
s2=at^2/2-a(t-2)^2/2=2at-2a
由题意可得：s1:s2=1：2=2a：（2at-2a）
s2-s1=10=2at-2a-2a=2at-4a
故有：2a=at-a，
at-2a=5
解得：t=3s，a=5m/s^2
则：L=at^2/2=5*9/2=22.5m
绛旓細2鈭2m/s 鍘熷洜锛歏^2=2aS+V涓璣2 V涓璣2=2aS 鍏朵腑鏂滈潰闀垮害=2S 鎵浠 V=2鈭2m/s
绛旓細浣犺繖涔堢畻锛氬厛姹鍔犻搴︼紝宸茬煡鐢遍潤姝㈠紑濮锛屽垯鍓嶄笁绉掔殑浣嶇Щs=0.5*a*(t鐨勫钩鏂癸級锛宻=4.5锛宼=3锛屽垯a=1锛屽悗涓夌s=vt+0.5*a*(t鐨勫钩鏂)锛宻=10.5锛宼=3锛宎=1锛屽垯v=2銆傚垯鍚庝笁绉掔殑璧峰閫熷害涓2锛岀敤鍏紡v=at(鍥犱负鍒濆涓洪潤姝)锛屽垯t=2锛屾墍浠ユ眰鐨勬绘椂闂翠负3+2=5锛屽垯鎬讳綅绉讳负s=o.5*a*(...
绛旓細璁撅紝鍔犻熷害涓篴,鐗╀綋鍦鍊掓暟绗笁绉掔殑鐬椂閫熷害涓篤o 鐢遍鎰忥紝鍙緱锛歋2-S1=6 S1:S2=3:7 鑱旂珛涓ゅ紡锛氬彲寰,S1=4.5, S2=10.5 鐗╀綋婊戝姩鐨勬渶鍒濅笁绉掑唴鏈夛細s1=1/2at锛堝钩鏂癸級鍙緱锛歛=1 鐗╀綋婊戝姩鏈鍚庝笁绉掑唴鏈夛細S2=Vot+1/2at锛堝钩鏂癸級鍙緱鍑 Vo 锛堟垜涓嶇畻浜嗏︹︼級鍥犱负鐗╀綋涓鍖鍔犻杩愬姩锛氭墍浠 ...
绛旓細锛1锛夊浜庡垵閫熷害涓洪浂鐨鍖鍔犻鐩寸嚎杩愬姩锛屽湪杩炵画鐩哥瓑鏃堕棿鍐呯殑浣嶇Щ涔嬫瘮涓1锛3锛屽嵆x1锛歺2=1锛3锛寈2-x1=1.8m锛岃仈绔嬩袱寮忚В寰楋細x1=0.9m锛寈2=2.7m锛庯紙2锛夋牴鎹柍x=aT2锛孴=3s锛屽垯锛歛=x2?x1T2锛2.7?0.99锛0.2m/s2锛庯紙3锛鐗╀綋鍦绗3s鏈殑閫熷害绛変簬6s鍐呯殑骞冲潎閫熷害锛屽垯锛歷锛漻1+x22T锛0....
绛旓細鏂规硶涓銆佸墠3s鐨勫钩鍧囬熷害绛変簬鏃堕棿涓偣锛屽嵆1.5s鏃剁殑鐬椂閫熷害锛屼负0.9/3=0.3m/s a=0.3/1.5=0.2m/s^2 鐢变綅绉诲叕寮忓緱鏈鍚3绉掔殑鍒濋熷害涓簐1=0.4m/s锛屽啀鐢ㄩ熷害鍏紡寰楁湯閫熷害涓簐2=1.0m/s 鎬婚暱x=(v2)^2/(2*a)=2.5m 鏂规硶浜屻佽鍔犻搴︿负a锛屾绘椂闂翠负t锛鏂滈潰闀垮害涓簒.鏍规嵁杩愬姩瀛﹀叕寮忔湁...
绛旓細璁鐗╀綋鍒拌揪搴曠鐨勯熷害涓簐2锛屽埌杈句腑鐐规椂鐨勯熷害涓簐1锛屾牴鎹熷害浣嶇Щ鍏紡寰楋細v12锛2a?x2v22锛2ax鑱旂珛涓ゅ紡瑙ｅ緱锛歷2锛2v1锛22m/s锛庢晠D姝ｇ‘锛孉銆丅銆丆閿欒锛庢晠閫夛細D锛
绛旓細S1 S2 t1 t2 S1=1/2at12锛屽嵆4.5锛4.5a锛屾墍浠锛1m/s2 S2=v2t2+1/2at22 鍗10.5锛漹2+4.5锛屾墍浠2锛6m/s 璁炬渶鍚3s鍓嶅凡婊戣鏃堕棿涓篢锛屽垯 鐢变簬v2锛漚T锛屾墍浠锛6s 鎵浠ユ诲叡婊戣鏃堕棿涓9s 鎵浠 S=1/2a锛圱+t2锛2锛0.5*1*81锛40.5m ...
绛旓細璁鏂滈潰鎬婚暱2x.鏍规嵁鍒濋熷害涓洪浂鐨鍖鍔犻鐩寸嚎杩愬姩鍏紡锛歺=1/2at^2鍒楀紡瀛愭湁x=1/2a*1^2.2x=1/2aT^2.瑙ｅ緱T锛濇牴鍙2.
绛旓細1m/s 瑙ｆ瀽锛鍖鍔犻鐩寸嚎杩愬姩涓紝閫熷害涓庝綅绉荤殑鍏崇郴涓 v^2-v0^2=2ax (v涓烘湯閫熷害锛寁o涓哄垵閫熷害锛夊洜涓哄師鏉闈欐锛屾墍浠0=0,鎵浠 v^2=2ax (1)璁捐繃涓偣鏃堕熷害涓簐1锛屽垯姝ゆ椂浣嶇Щ涓0.5x v1^2=2a*0.5x=ax 锛2锛夎仈绯伙紙1锛夊紡鍜岋紙2锛夊紡鍙眰寰梫1=1m/s ...
绛旓細鍦ㄦ枩闈涓婂仛鍖鍔犻锛屽垯鐢眘1=(v0+vt)/2*t寰梥=(0+v1)*3/2=1.5v1 鍦ㄦ按骞抽潰涓婂仛鍖鍑忛燂紝骞朵笖鍖鍑忛熺殑鍒濋熷害澶у皬绛変簬鏂滈潰涓婂寑鍔犻熺殑鏈熷害锛屽洜姝ゅ彲寰梥2=(v0+vt)/2*t=(v1+0)*9/2=4.5v1 鎵浠1:s2=1:3
