Bài 5: Tốc độ và vận tốc - Bài tập số 3 (nâng cao)

Home Lớp 10 Vật lý 10 - Sách kết nối tri thức Bài 5: Tốc độ và vận tốcBài tập nâng cao

{"common":{"save":0,"post_id":"6313","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5806","mon_id":"1158790","chapter_id":"1158795","post_id":"6313","question":"M\u1ed9t ng\u01b0\u1eddi \u0111i t\u1eeb A \u0111\u1ebfn B v\u1edbi t\u1ed1c \u0111\u1ed9 12 km\/h \u0111\u01b0\u1ee3c quãng \u0111\u01b0\u1eddng s1 thì xe h\u01b0, ph\u1ea3i s\u1eeda ch\u1eefa m\u1ea5t 15 phút. Trong quãng \u0111\u01b0\u1eddng còn l\u1ea1i $s_2$<\/span> ng\u01b0\u1eddi \u0111ó \u0111i v\u1edbi t\u1ed1c \u0111\u1ed9 15 km\/h thì \u0111\u1ebfn n\u01a1i v\u1eabn s\u1edbm h\u01a1n d\u1ef1 \u0111\u1ecbnh 30 phút. Tìm quãng \u0111\u01b0\u1eddng $s_1$<\/span>, bi\u1ebft quãng \u0111\u01b0\u1eddng AB dài 60 km, th\u1eddi gian \u0111i t\u1eeb A \u0111\u1ebfn B m\u1ea5t 5 gi\u1edd.<\/p>","options":["A. 45 km","B. 30 km","C. 25 km","D. 40 km"],"correct":"1","level":"3","hint":"Tìm quãng \u0111\u01b0\u1eddng b\u01a1i \u0111\u01b0\u1ee3c và \u0111\u1ed9 d\u1ecbch chuy\u1ec3n c\u1ee7a ng\u01b0\u1eddi \u0111ó, t\u1eeb \u0111ó tính \u0111\u01b0\u1ee3c t\u1ed1c \u0111\u1ed9 trung bình và v\u1eadn t\u1ed1c trung bình.$t_1=\\frac{s_1}{\\upsilon_1}$<\/span><\/p>","answer":"Ch\u1ecdn A. G\u1ecdi $t_1$<\/span> là th\u1eddi gian ng\u01b0\u1eddi \u0111ó \u0111i v\u1edbi t\u1ed1c \u0111\u1ed9 12 km\/h, ta có:   $t_2$<\/span> là th\u1eddi gian ngh\u1ec9 15 phút \u0111\u1ec3 s\u1eeda xe: $t_2=15$<\/span> phút = 0,25 gi\u1edd $t_3$<\/span> là th\u1eddi gian ng\u01b0\u1eddi \u0111ó \u0111i v\u1edbi t\u1ed1c \u0111\u1ed9 15 km\/h, ta có:  $t_3=\\frac{s_2}{\\upsilon_2}=\\frac{AB-s_1}{\\upsilon_2}$<\/span> t\u1ed1c \u0111\u1ed9 trung bình trên c\u1ea3 quãng \u0111\u01b0\u1eddng là:<\/strong><\/span><\/p>$\\upsilon= \\frac{AB}{t}=\\frac{AB}{t_1+t_2+t_3}=\\frac{AB}{\\frac{s_1}{\\upsilon_1}+t_2+\\frac{AB-s_1}{\\upsilon_2}}$<\/span><\/strong><\/span><\/p>$\\Leftrightarrow \\frac{60}{5}=\\frac{60}{\\frac{s_1}{12}+0,25+\\frac{60-s_1}{15}}$<\/span><\/strong><\/span><\/p>$\\Rightarrow 5= \\frac{s_1}{12}+0,25+\\frac{60-s_1}{15}$<\/span><\/strong><\/span><\/p>$\\Rightarrow s_1=45(km)$<\/span><\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2022-10-17 08:52:08","option_type":"txt","len":0},{"id":"5807","mon_id":"1158790","chapter_id":"1158795","post_id":"6313","question":"M\u1ed9t ng\u01b0\u1eddi \u0111i t\u1eeb A t\u1edbi B v\u1edbi t\u1ed1c \u0111\u1ed9 12 km\/h. N\u1ebfu ng\u01b0\u1eddi \u0111ó t\u0103ng t\u1ed1c thêm 3 km\/h thì \u0111\u1ebfn n\u01a1i s\u1edbm h\u01a1n 1 gi\u1edd. Tìm quãng \u0111\u01b0\u1eddng AB và th\u1eddi gian d\u1ef1 \u0111\u1ecbnh \u0111i t\u1eeb A t\u1edbi B.<\/p>","options":["A. 60 (km) – 5 (h)","B. 35 (km) – 2 (h)","C. 40 (km) – 1 (h)","D. 80 (km) – 2,5 (h)"],"correct":"1","level":"3","hint":"G\u1ecdi $t_1$<\/span> là th\u1eddi gian ng\u01b0\u1eddi \u0111ó \u0111i t\u1eeb A \u0111\u1ebfn B v\u1edbi t\u1ed1c \u0111\u1ed9 $\\upsilon_1$<\/span> , $t_2$<\/span> là th\u1eddi gian ng\u01b0\u1eddi \u0111ó \u0111i t\u1eeb A \u0111\u1ebfn B v\u1edbi t\u1ed1c \u0111\u1ed9 $\\upsilon_2$<\/span> . N\u1ebfu ng\u01b0\u1eddi \u0111ó t\u0103ng t\u1ed1c thêm 3 km\/h thì \u0111\u1ebfn n\u01a1i s\u1edbm h\u01a1n 1 gi\u1edd => $t_2 = t_1 \u2013 1 (*)$<\/span> Gi\u1ea3i ph\u01b0\u01a1ng trình $(*)$<\/span> \u0111\u1ec3 tìm quãng \u0111\u01b0\u1eddng AB và th\u1eddi gian d\u1ef1 \u0111\u1ecbnh \u0111i t\u1eeb A \u0111\u1ebfn B<\/p>","answer":"Ch\u1ecdn A. G\u1ecdi s là \u0111\u1ed9 dài quãng \u0111\u01b0\u1eddng AB; $t_1$<\/span> là th\u1eddi gian ng\u01b0\u1eddi \u0111ó \u0111i t\u1eeb A \u0111\u1ebfn B v\u1edbi t\u1ed1c \u0111\u1ed9 $\\upsilon_1 =12(km\/h)$<\/span> , $t_2$<\/span> là th\u1eddi gian ng\u01b0\u1eddi \u0111ó \u0111i t\u1eeb A \u0111\u1ebfn B v\u1edbi t\u1ed1c \u0111\u1ed9 $\\upsilon_2$<\/span> . Ta có:  $ \\upsilon_2=\\upsilon_1+3=12+3=15(km\/h)$<\/span> Khi ng\u01b0\u1eddi \u0111ó \u0111i v\u1edbi t\u1ed1c \u0111\u1ed9  $ \\upsilon_2$<\/span> thì \u0111\u1ebfn n\u01a1i s\u1edbm h\u01a1n 1 gi\u1edd so v\u1edbi khi \u0111i v\u1edbi t\u1ed1c \u0111\u1ed9  $ \\upsilon_1$<\/span>  $\\Rightarrow t_2=t_1-1$<\/span><\/strong><\/span><\/p>$\\Leftrightarrow \\frac{s}{\\upsilon_2}=\\frac{s}{\\upsilon_1}-1$<\/span><\/strong><\/span><\/p>$\\Leftrightarrow \\frac{s}{15}=\\frac{s}{12}-1$<\/span><\/strong><\/span><\/p>$\\Rightarrow s= 60(km\/h)$<\/span> Th\u1eddi gian d\u1ef1 \u0111\u1ecbnh \u0111i t\u1eeb A t\u1edbi B là:  $t_1=\\frac{s}{\\upsilon_1}=\\frac{60}{12}=5(h)$<\/span><\/strong><\/span><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2022-10-17 09:04:39","option_type":"txt","len":2},{"id":"5808","mon_id":"1158790","chapter_id":"1158795","post_id":"6313","question":"M\u1ed9t xe máy \u0111i n\u1eeda \u0111o\u1ea1n \u0111\u01b0\u1eddng \u0111\u1ea7u tiên v\u1edbi t\u1ed1c \u0111\u1ed9  $\\upsilon_1$<\/span> = 60 km\/h và n\u1eeda \u0111o\u1ea1n \u0111\u01b0\u1eddng sau v\u1edbi t\u1ed1c \u0111\u1ed9  $\\upsilon_2$<\/span> = 40 km\/h. Tính t\u1ed1c \u0111\u1ed9 trung bình trên c\u1ea3 \u0111o\u1ea1n \u0111\u01b0\u1eddng.<\/p>","options":["A. 100 km\/h","B. 60 km\/h","C. 24 km\/h","D. 48 km\/h"],"correct":"4","level":"3","hint":"Vi\u1ebft công th\u1ee9c tính t\u1ed1c \u0111\u1ed9, t\u1eeb \u0111ó rút th\u1eddi gian \u0111i \u0111\u01b0\u1ee3c trên m\u1ed7i \u0111o\u1ea1n \u0111\u01b0\u1eddng. Thay th\u1eddi gian \u0111i \u0111\u01b0\u1ee3c trên m\u1ed7i \u0111o\u1ea1n \u0111\u01b0\u1eddng vào công th\u1ee9c tính \u0111\u01b0\u1ee3c t\u1ed1c \u0111\u1ed9 trung bình trên c\u1ea3 \u0111o\u1ea1n \u0111\u01b0\u1eddng.<\/p>","answer":"Ch\u1ecdn D. G\u1ecdi s là \u0111\u1ed9 dài quãng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c; $s_1-t_1$<\/span>  l\u1ea7n l\u01b0\u1ee3t là quãng \u0111\u01b0\u1eddng và th\u1eddi gian xe máy \u0111i v\u1edbi t\u1ed1c \u0111\u1ed9 $\\upsilon_1$<\/span> , $s_2-t_2$<\/span>  l\u1ea7n l\u01b0\u1ee3t là quãng \u0111\u01b0\u1eddng và th\u1eddi gian xe máy \u0111i v\u1edbi t\u1ed1c \u0111\u1ed9 $\\upsilon_2$<\/span> . Ta có:  $\\upsilon_1=\\frac{s_1}{t_1}=\\frac{\\frac{s}{2}}{t_1}=\\frac{s}{2.t_1}\\Rightarrow t_1=\\frac{s}{2.\\upsilon_1}$<\/span><\/strong><\/span><\/p>$\\upsilon_2=\\frac{s_2}{t_2}=\\frac{\\frac{s}{2}}{t_2}=\\frac{s}{2.t_2}\\Rightarrow t_2=\\frac{s}{2.\\upsilon_2}$<\/span> V\u1eadn t\u1ed1c trung bình trên c\u1ea3 quãng \u0111\u01b0\u1eddng là:  $\\upsilon=\\frac{s_1+s_2}{t_1+t_2}=\\frac{\\frac{s}{2}+\\frac{s}{2}}{\\frac{s}{2.\\upsilon_1}+\\frac{s}{2.\\upsilon_2}}=\\frac{1}{\\frac{1}{2.\\upsilon_1}+\\frac{1}{2.\\upsilon_2}}=\\frac{1}{\\frac{1}{2.60}+\\frac{1}{2.40}}=48(km\/h)$<\/span> V\u1eady t\u1ed1c \u0111\u1ed9 trung bình trên c\u1ea3 \u0111o\u1ea1n \u0111\u01b0\u1eddng là 48 km\/h.<\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2022-10-17 09:21:11","option_type":"txt","len":1}]}