一辆车沿平直公路从甲站开往乙站，启动过程可简化为加速度两米每秒的平方的匀加速直线运动，加速行驶五秒
\u4e00\u8f86\u6c7d\u8f66\u6cbf\u5e73\u76f4\u516c\u8def\u4ece\u7532\u7ad9\u5f00\u5f80\u4e59\u7ad9\uff0c\u8d77\u52a8\u65f6\u52a0\u901f\u5ea6\u4e3a2m/S\u5e73\u65b9\uff0c\u52a0\u901f\u884c\u9a765s\u540e\uff0c\u5747\u5300\u884c\u9a762min\uff0c\u7136\u540e\u8bf7\u95ee\u9898\u76ee\u4e2d\u6709\u6ca1\u6709\u7ed9\u51fa\u5239\u8f66\u4ee5\u540e\u7684\u52a0\u901f\u5ea6\uff1f
\u9996\u5148\u7b97\u7532\u4e59\u4e24\u7ad9\u7684\u8def\u7a0b
\u5f00\u59cb\u7684\u65f6\u5019\u662f\u521d\u901f\u5ea6\u4e3a0\u7684\u5300\u52a0\u901f\u8fd0\u52a8 s1=1/2at^2 s1=1/2*2*5^2=25m
\u5300\u901f\u884c\u9a76\u4e86\u4e24\u5206\u949f \u5300\u901f\u8fd0\u52a8\u65f6\u901f\u5ea6v=at v=2*5=10m/s
\u5300\u901f\u884c\u9a76\u7684\u8def\u7a0b s2= v*t=10*2*60=1200m
\u603b\u8def\u7a0bs=s1+s2+50=1275m
\u5239\u8f66\u540e\u6c7d\u8f66\u505a\u5300\u51cf\u901f\u8fd0\u52a8\uff0c\u7f3a\u5c11\u6761\u4ef6\uff0c\u6682\u65f6\u65e0\u6cd5\u89e3\u7b54\u3002\u3002\u3002
\u9996\u5148\u8981\u5206\u6790\u9898\u76ee\uff1a\u672c\u9898\u6709\u4e09\u4e2a\u9636\u6bb5\uff0c\u7b2c\u4e00\uff0c\u52a0\u901f\u5ea6\u4e3a2m/s^2\u7684\u5300\u52a0\u901f\u76f4\u7ebf\u8fd0\u52a8\u8fd0\u52a8\uff0c\u7b2c\u4e8c\uff0c\u662f\u901f\u5ea6\u4e3aV1\u7684\u5300\u901f\u76f4\u7ebf\u8fd0\u52a8\uff0c\u7b2c\u4e09\uff0c\u662f\u521d\u901f\u5ea6\u4e3aV1\u7684\u5300\u51cf\u901f\u76f4\u7ebf\u8fd0\u52a8\u3002
\u5300\u52a0\u901f\u76f4\u7ebf\u8fd0\u52a8\uff1at=5s \uff1ba=2m/s^2 ,V1=at=2*5=10m/s\uff0cS1=0.5*a*t^2=25m
\u5300\u901f\u76f4\u7ebf\u8fd0\u52a8:t=2min=120s, S2=V*t=V1*t=10*120=1200m
\u5300\u51cf\u901f\u76f4\u7ebf\u8fd0\u52a8:Vt=0,V0=10m/s,S3=50, (Vt)^2-(V0)^2=-2as \u56e0\u4e3a\u662f\u51cf\u901f\u8fd0\u52a8\uff0c\u6240\u4ee5a\u53d6\u8d1f\u503c
0-10^2=-2a*50 a=1m/s^2 \uff0c \u56e0\u4e3a Vt=V0-at \uff0c0=10-at \uff0c t=10s
S\u603b=S1+S2+S3=25+1200+50=1275m
t\u603b\u5c31\u662f\u4e09\u4e2a\u9636\u6bb5\u6240\u7528\u7684\u65f6\u95f4\u603b\u548c\uff1a\u5373t\u603b=5+120+10=135s
\u5168\u8fc7\u7a0b\u7684\u5e73\u5747\u901f\u5ea6V\u5e73=S\u603b/t\u603b=1275/135=85/9\uff08m/s\uff09
2、刹车过程初速度为V1＝10m/s，末速度为0，则平均速度V3＝V1/2＝5m/s，运动时间t3=S3/V3=50/5=10s。
3、S1＝1/2 at1^2=1/2 * 2 *5^2=25m，S2=V1 * 15=10*15=150m，
总位移S＝S1+S2+S3＝25+150+50＝225m，总时间t=t1+t2+t3=5+15+10=30s，
平均速度V＝S/t=225/30=7.5m/s。
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绛旓細绛旓細鐢层佷箼涓ょ珯鐨勮窛绂讳负1275m锛屾苯杞︿粠鐢茬珯鍒涔欑珯鎵鐢ㄦ椂闂翠负135s
绛旓細鍖鍔犻熺洿绾胯繍鍔細t=5s 锛沘=2m/s^2 ,V1=at=2*5=10m/s锛孲1=0.5*a*t^2=25m 鍖閫熺洿绾胯繍鍔:t=2min=120s, S2=V*t=V1*t=10*120=1200m 鍖鍑忛熺洿绾胯繍鍔:Vt=0,V0=10m/s,S3=50, (Vt)^2-(V0)^2=-2as 鍥犱负鏄噺閫熻繍鍔紝鎵浠鍙栬礋鍊 0-10^2=-2a*50 a=1m/s...
绛旓細璧峰姩闃舵琛岄┒浣嶇Щ涓猴細S 1 = 1 2 at 1 2 = 1 2 脳2脳25m=25m 鍖閫熻椹剁殑閫熷害涓猴細V=at 1 =10m/s鍖閫熻椹剁殑浣嶇Щ涓猴細S 2 =Vt 2 =10脳120m=1200m鎵浠鐢茬珯鍒涔欑珯涔嬮棿鐨勮窛绂讳负25+1200+50m=1275m鏁呯瓟妗堜负锛1275 ...
绛旓細鍖閫熺殑浣嶇Щx2=vt2=10脳120m=1200m锛屽垯鐢蹭箼涓ょ珯鐨勮窛绂粁=x1+x2+x3=25+1200+50m=1275m锛庢牴鎹畑3锛漹2t3寰楋紝鍖鍑忛熻繍鍔ㄧ殑鏃堕棿t3锛2x3v锛2脳5010s锛10s锛屽垯姹借溅浠庣敳绔鍒涔欑珯鐨勬椂闂磘=t1+t2+t3=5+120+10s=135s锛庣瓟锛氱敳銆佷箼涓ょ珯鐨勮窛绂讳负1275m锛屾苯杞︿粠鐢茬珯鍒颁箼绔欑殑鏃堕棿涓135s锛
绛旓細璧峰姩闃舵琛岄┒浣嶇Щ涓猴細S 1 = 1 2 a t 1 2 鍖閫熻椹剁殑閫熷害涓猴細V=at 1 鍖閫熻椹剁殑浣嶇Щ涓猴細S 2 =Vt 2 鍒硅溅娈电殑鏃堕棿涓猴細S 3 = V 2 t 3 姹借溅浠庣敳绔鍒涔欑珯鐨勫钩鍧囬熷害涓猴細V= S 1 + S 2 + S 3 t 1 + t 2 + t...
绛旓細1 銆 涓杈嗘苯杞︽部骞崇洿鍏矾浠庣敳鎴寮寰涔欑珯,鍚姩鍔犻熷害涓2绫虫瘡浜屾鏂瑰锛屽姞閫熷害褰㈠紡5绉掞紝鍚庡寑閫熷舰寮2鍒嗛挓锛岀劧鍚庡埞杞︼紝婊戣50绫筹紝姝ｅソ鍒拌揪涔欑珯锛屾眰姹借溅浠庣敳鎴樺埌涔欑珯鐨勫钩鍧囬熷害銆傚姞閫燂細S1=1/2at^2=1/2*2*25=25m,v1=at=2*5=10m/s 鍖閫熻繍鍔細s2=v1t2=10*2*60=1200m 鍒硅溅锛歵3=s3/(v1...
绛旓細瑙ｏ細姹借溅鍚姩闃舵琛岄┒浣嶇Щ涓猴細 鍖閫熻椹剁殑閫熷害涓猴細v=at 1 鍖閫熻椹剁殑浣嶇Щ涓猴細x 2 =vt 2 鍒硅溅闃舵鐨勪綅绉讳负锛 鑱旂珛鍥涘紡鍙緱姹借溅浠庣敳绔鍒涔欑珯鐨勫钩鍧囬熷害涓猴細 =
绛旓細涓嬭浇浣滀笟甯～閭璇风爜4502526瀛﹂湼甯綘鍒嗗垎閽
绛旓細杩欓鍙互鐢ㄤ綔鍥炬硶锛屾妸鍔犻熷拰鍖閫熼樁娈电殑V-t鍥句綔鍑猴紝鍙互姹傚緱姹借溅鍔犻熷拰鍖閫熼樁娈佃椹剁殑浣嶇Щ涓篨1=1/2*2*5^2+2*5*2*60=1225m,鍔犱笂婊戣浣嶇Щ锛屾讳綅绉讳负x=1275m銆傚埞杞︽粦琛屾椂闂村彲鐢50绫抽櫎浠ユ粦琛屽钩鍧囬熷害=10s.鎵浠姹借溅浠庣敳鍒涔骞冲潎閫熷害涓簐=1275/(10+5+120)=9.4444...m/s ...
