Isibonelo sombuzo wengxoxo mayelana nokwandisa isivinini
Ijubane liwumqondo oyisisekelo ku-physics, ikakhulukazi ku-kinematics, igatsha le-physics elifunda ukunyakaza kwezinto ngaphandle kokucabangela imbangela yalokho kunyakaza. Lo mqondo awubalulekile kuphela kwezemfundo kodwa nasezinhlelweni zansuku zonke ezifana nobunjiniyela, ezemidlalo, kanye nezokuthutha. Lesi sihloko sizoxoxa ngezinkinga eziningana zezibonelo ezihlobene nejubane, eziphelele ngezincazelo zesinyathelo ngesinyathelo ukusiza ukuqonda.
Umqondo Oyisisekelo Wokwengeza Isivinini
Ngaphambi kokungena emibuzweni eyisibonelo, kungaba kuhle ngathi ukukhumbula imiqondo eyisisekelo mayelana nesivinini kanye nokusheshisa.
1. Ijubane (v) lichazwa njengoshintsho lwesikhundla ngesikhathi seyunithi ngayinye.
2. Ukusheshisa (a) inani elichaza ushintsho lwejubane ngeyunithi yesikhathi.
Ifomula eyisisekelo yokusheshisa yile:
\[ a = \frac{\Delta v}{\Delta t} \]
Kuphi:
– \( a \) ukusheshisa,
– \( \Delta v \) ushintsho lwejubane,
– \( \Delta t \) yisikhathi esiphakathi.
Isibonelo Umbuzo 1
Umbuzo:
Imoto iqala ihambe ngesivinini esingama-10 m/s. Ngemva kwemizuzwana emi-5, isivinini semoto siba ngama-20 m/s. Iyini isilinganiso sokusheshisa semoto?
Ingxoxo:
Kuyaziwa:
– Ijubane lokuqala (\( v_0 \)) = 10 m/s,
– Ijubane lokugcina (\( v_f \)) = 20 m/s,
– Isikhathi (\( \Delta t \)) = 5 s.
Singasebenzisa ifomula yokusheshisa:
\[ a = \frac{v_f – v_0}{\Delta t} \]
Faka amanani aziwayo esikhundleni sawo:
\[ a = \frac{20 – 10}{5} \]
\[ a = \frac{10}{5} \]
\[ a = 2 \, \text{m/s}^2 \]
Ngakho-ke, ukusheshisa okujwayelekile kwemoto kungama-2 m/s².
Isibonelo Umbuzo 2
Umbuzo:
Isitimela siqala siphumule futhi sifinyelela isivinini esingama-30 m/s ngemizuzwana eyi-10. Bala ukusheshisa kwaso okujwayelekile kanye nebanga elihanjiwe phakathi naleso sikhathi.
Ingxoxo:
Ukusheshisa:
Kuyaziwa:
– Ijubane lokuqala (\( v_0 \)) = 0 m/s (ngoba liqala esimweni esingashintshi),
– Ijubane lokugcina (\( v_f \)) = 30 m/s,
– Isikhathi (\( \Delta t \)) = 10 s.
Ukusebenzisa ifomula yokusheshisa:
\[ a = \frac{v_f – v_0}{\Delta t} = \frac{30 – 0}{10} = 3 \, \text{m/s}^2 \]
Ibanga elihanjiwe:
Ukuze sibale ibanga (\( s \)) elihanjiwe, singasebenzisa esinye sezibalo ze-kinematic:
\[ s = v_0 t + \frac{1}{2} at^2 \]
Faka amanani aziwayo esikhundleni sawo:
\[ s = 0 \cdot 10 + \frac{1}{2} \cdot 3 \cdot (10)^2 \]
\[ s = \frac{1}{2} \cdot 3 \cdot 100 \]
\[ s = 150 \, \umbhalo{m} \]
Ngakho-ke, ibanga elihanjwa yisitimela lingamamitha ayi-150.
Isibonelo Umbuzo 3
Umbuzo:
Isithuthuthu simi endaweni ethile futhi siqala ukuhamba ngesivinini esingaguquki esingu-4 m/s². Ngemva kokuhamba imizuzwana engu-8, ingakanani isivinini sokugcina kanye nebanga elihanjwa yilesi sithuthuthu?
Ingxoxo:
Isivinini sokugcina:
Kuyaziwa:
– Ijubane lokuqala (\( v_0 \)) = 0 m/s (ngoba liqala esimweni esingashintshi),
– Ukusheshisa (\( a \)) = 4 m/s²,
– Isikhathi (\( t \)) = 8 s.
Ukusebenzisa ifomula yesivinini:
\[ v_f = v_0 + ku- \]
Faka amanani aziwayo esikhundleni sawo:
\[ v_f = 0 + 4 \cdot 8 \]
\[ v_f = 32 \, \umbhalo{m/s} \]
Ibanga elihanjiwe:
Ukusebenzisa ifomula yebanga:
\[ s = v_0 t + \frac{1}{2} at^2 \]
Faka amanani aziwayo esikhundleni sawo:
\[ s = 0 \cdot 8 + \frac{1}{2} \cdot 4 \cdot (8)^2 \]
\[ s = \frac{1}{2} \cdot 4 \cdot 64 \]
\[ s = 128 \, \umbhalo{m} \]
Ngakho-ke, ngemva kokuhamba imizuzwana engu-8, isivinini sokugcina sesithuthuthu singama-32 m/s kanti ibanga esilihambile lingamamitha angu-128.
Isibonelo Umbuzo 4
Umbuzo:
Ibhola liphonswa phezulu phezulu ngesivinini sokuqala esingu-20 m/s. Ngemva kokufinyelela iphuzu lalo eliphakeme kakhulu, ibhola liphinde liwele phansi ngesivinini ngenxa yamandla adonsela phansi \( g = 9.8 \, \text{m/s}^2 \). Kuthatha isikhathi esingakanani ukuthi ibhola liphinde lifike phansi?
Ingxoxo:
Isikhathi esithathayo ukukhuphuka nokwehla siyafana. Ngakho-ke sidinga nje ukubala isikhathi esithathayo ukukhuphuka, bese siphindaphinda ngo-2 ukuze sithole isikhathi esiphelele.
Kuyaziwa:
– Ijubane lokuqala (\( v_0 \)) = 20 m/s,
– Isivinini esisezingeni eliphezulu kakhulu (\( v_f \)) = 0 m/s (ngoba sima okwesikhashana),
– Ukusheshisa ngenxa yamandla adonsela phansi (\(g \)) = 9.8 m/s².
Ukusebenzisa ifomula yesivinini:
\[ v_f = v_0 + (-g) t \]
Faka amanani aziwayo esikhundleni sawo:
\[ 0 = 20 – 9.8 t \]
\[ 9.8 t = 20 \]
\[ t = \frac{20}{9.8} \]
\[t \cishe 2.04 \, \umbhalo{s} \]
Lesi yisikhathi sokuba ibhola lifinyelele iphuzu lalo eliphezulu. Ngakho-ke, isikhathi esiphelele sokukhuphuka nokuwa yilesi:
\[ 2 \cdot 2.04 \cishe 4.08 \, \text{s} \]
Ngakho-ke, isikhathi esiphelele esithathwayo ukuze ibhola liphinde lifike phansi singamasekhondi angu-4.08.
Isiphetho
Kuzo zonke izinkinga okuxoxwe ngazo ngenhla, isinyathelo esiyinhloko ukuqonda imiqondo eyisisekelo yejubane kanye nokusheshisa kanye nendlela esetshenziswa ngayo kumafomula athile. Nakuba izinkinga zihluka, indlela inamathela ezimisweni eziyisisekelo zefiziksi. Kuthenjwa ukuthi ngokwenza lezi zinkinga, abafundi bazothola ukuqonda okujulile kokuthi ijubane kanye nokusheshisa kuhlangana kanjani ekunyakazeni kwezinto.
Yiqiniso, ekusetshenzisweni kwansuku zonke, ukuqonda lo mqondo kungaba usizo kakhulu, hhayi nje kuphela kwezemfundo kodwa nasemikhakheni ehlukahlukene yobungcweti njengobunjiniyela, ezokuthutha, neminye. Khumbula njalo ukuqonda inkinga kuqala, ngaphambi kokuzama ukuyixazulula, ukuze inqubo yokuqonda nokuxazulula inkinga ibe lula futhi isebenze kakhudlwana.
