Chuyển động biến đổi thẳng đều

Home Lớp 10 Vật lý 10 - Sách kết nối tri thức Bài 9: Chuyển động biến đổi thẳng đềuBài tập nâng cao

{"common":{"save":0,"post_id":"6345","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5896","mon_id":"1158790","chapter_id":"1158795","post_id":"6345","question":"M\u1ed9t ch\u1ea5t \u0111i\u1ec3m chuy\u1ec3n \u0111\u1ed9ng \u0111\u01b0\u1ee3c mô t\u1ea3 qua \u0111\u1ed3 th\u1ecb v – t cho d\u01b0\u1edbi \u0111ây. Hãy tính v\u1eadn t\u1ed1c c\u1ee7a ch\u1ea5t \u0111i\u1ec3m t\u1ea1i t = 8 s?<\/p><svg height=\"272\" width=\"324\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g><title><\/title><rect fill=\"#fff\" height=\"274\" id=\"canvas_background\" width=\"326\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" width=\"100%\" x=\"0\" y=\"0\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g><title><\/title><image height=\"272\" id=\"svg_1\" width=\"324\" x=\"0\" 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y=\"0\"><\/image> <\/g> <\/svg><\/span><\/p>","options":["A. 10 m\/s","B. 8 m\/s","C. 9 m\/s","D. 11 m\/s"],"correct":"3","level":"3","hint":"T\u1eeb \u0111\u1ed3 th\u1ecb, vi\u1ebft ph\u01b0\u01a1ng trình v\u1eadn t\u1ed1c (ph\u01b0\u01a1ng trình v\u1eadn t\u1ed1c có d\u1ea1ng $\\nu=\\nu_0+at$<\/span>) Thay t = 8 (s) vào ph\u01b0\u01a1ng trình v\u1eadn t\u1ed1c \u0111\u1ec3 tìm v.<\/p>","answer":"Ch\u1ecdn C.<\/strong><\/span> Ta có: $\\begin{cases}\\nu_0=1m\/s\\\\a=\\frac{\\triangle \\nu}{\\triangle t}=\\frac{2-1}{1}=1m\/s^2\\end{cases}\\Rightarrow \\nu_t=\\nu_0+at=1+t$<\/span>  T\u1ea1i t = 8 s  $\\Rightarrow \\nu_t=1+8=9(m\/s)$<\/span> V\u1eady v\u1eadn t\u1ed1c c\u1ee7a ch\u1ea5t \u0111i\u1ec3m t\u1ea1i t = 8 s là 9 m\/s.  <\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2022-10-26 08:42:26","option_type":"txt","len":0},{"id":"5897","mon_id":"1158790","chapter_id":"1158795","post_id":"6345","question":"M\u1ed9t v\u1eadt chuy\u1ec3n \u0111\u1ed9ng bi\u1ebfn \u0111\u1ed5i \u0111\u1ec1u có ph\u01b0\u01a1ng trình \u0111\u1ed9 d\u1ecbch chuy\u1ec3n là: $ d = 4t^2 + 20t $<\/span> (d \u0111o b\u1eb1ng m, t \u0111o b\u1eb1ng s). Tính v\u1eadn t\u1ed1c c\u1ee7a v\u1eadt lúc $ t = 3s$<\/span>.<\/p>","options":["A. 44 m\/s","B. 52 m\/s \n ","C. 22,5 m\/s","D. 39 m\/s"],"correct":"1","level":"3","hint":"T\u1eeb ph\u01b0\u01a1ng trình trên, suy ra gia t\u1ed1c và v\u1eadn t\u1ed1c ban \u0111\u1ea7u c\u1ee7a v\u1eadt V\u1eadn d\u1ee5ng công th\u1ee9c $\\nu_t=\\nu_0+at$<\/span> \u0111\u1ec3 tìm v\u1eadn t\u1ed1c c\u1ee7a v\u1eadt lúc $t = 3s.$<\/span><\/p>","answer":"Ch\u1ecdn A.<\/strong><\/span>  $ d = 4t^2 + 20t \\Rightarrow\\begin{cases}4=\\frac{1}{2}a\\\\20=\\nu_0\\end{cases}\\Rightarrow \\begin{cases}a=8(m\/s^2)\\\\\\nu_0=20(m\/s)\\end{cases}$<\/span> Ta có:  $\\nu_{t=3s}=\\nu_0+a.t=20+8.3=44(m\/s)$<\/span> V\u1eady v\u1eadn t\u1ed1c c\u1ee7a v\u1eadt lúc $t = 3s$<\/span> là 44 m\/s.<\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2022-10-26 08:48:11","option_type":"txt","len":1},{"id":"5898","mon_id":"1158790","chapter_id":"1158795","post_id":"6345","question":"M\u1ed9t ng\u01b0\u1eddi \u0111i xe máy \u0111ang \u0111i v\u1edbi v\u1eadn t\u1ed1c 36 km\/h thì th\u1ea5y tín hi\u1ec7u \u0111èn \u0111\u1ecf, ng\u01b0\u1eddi \u0111ó hãm phanh và d\u1eebng xe l\u1ea1i sau 8 s. Tính v\u1eadn t\u1ed1c trung bình c\u1ee7a ng\u01b0\u1eddi \u0111ó k\u1ec3 t\u1eeb khi hãm phanh \u0111\u1ebfn khi d\u1eebng l\u1ea1i.<\/p>","options":["A. 8,25 m\/s","B. 5 m\/s","C. 7,75 m\/s","D. 10 m\/s"],"correct":"2","level":"3","hint":"Vi\u1ebft ph\u01b0\u01a1ng trình \u0111\u1ed9 d\u1ecbch chuy\u1ec3n, tính \u0111\u1ed9 d\u1ecbch chuy\u1ec3n c\u1ee7a ng\u01b0\u1eddi \u0111ó sau 8 s. V\u1eadn t\u1ed1c trung bình  ${\\nu=}{d\\over t}$<\/span><\/p>","answer":"Ch\u1ecdn B.<\/strong><\/span> \u0110\u1ed5i 36 km\/h = 10 m\/s Ch\u1ecdn chi\u1ec1u d\u01b0\u01a1ng là chi\u1ec1u chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a ng\u01b0\u1eddi \u0111ó. Ta có:  $\\begin{cases}\\nu_0=1m\/s\\\\a=\\frac{\\triangle \\nu}{\\triangle t}=\\frac{0-10}{8}=-1,25m\/s^2\\end{cases}\\Rightarrow d=10t+\\frac{1}{2}(-1,25)t^2=10t-0.625t^2$<\/span> T\u1ea1i t = 8 (s), ta có  $d=10.8-0,625.8^2=40(m)$<\/span> V\u1eady v\u1eadn t\u1ed1c trung bình c\u1ee7a ng\u01b0\u1eddi \u0111ó k\u1ec3 t\u1eeb khi hãm phanh \u0111\u1ebfn khi d\u1eebng l\u1ea1i là:<\/p>$v_{tb}={d\\over t}={40\\over 8}=5(m\/s)$<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2022-10-26 08:53:45","option_type":"txt","len":1}]}