Dane: [latex]a=5frac{m}{s^2}[/latex] Obliczyć: [latex]s_{10}=? \v_{sr}=?[/latex] Rozwiązanie: [latex]s_{10}=s_c-s_9 \s_{10}=frac{1}{2}a{t_c}^2-frac{1}{2}a{t_{10}-1s}^2 \s_{10}=frac{1}{2}a({t_c}^2-{(t_{10}-1s)}^2) \s_{10}=frac{1}{2}*5frac{m}{s^2}((10s)^2-(10s-1s)^2) \s_{10}=frac{1}{2}*5frac{m}{s^2}((10s)^2-(9s)^2) \s_{10}=frac{5}{2}frac{m}{s^2}(100s^2-81s^2) \s_{10}=frac{5}{2}(100-81)m \s_{10}=frac{5}{2}*19m \s_{10}=frac{95}{2}m \s_{10}=47,5m[/latex] [latex]v_{sr}=frac{s_c}{t_c} \v_{sr}=frac{frac{1}{2}a{t_c}^2}{t_c} \v_{sr}=frac{1}{2}a{t_c} \v_{sr}=frac{1}{2}*5frac{m}{s^2}*10s \v_{sr}=frac{1}{2}*5frac{m}{s}*10 \v_{sr}=5frac{m}{s}*5 \v_{sr}=25frac{m}{s}[/latex]
Dane: [latex]a=5frac{m}{s^2}\ t=10s[/latex] Szukane: [latex]v_{sr}=?\ s=?[/latex] Rozwiązanie: [latex]s=s_{10}-s_{9}[/latex] gdzie [latex]s_{10}[/latex] - droga przebyta po 10 s, [latex]s_9[/latex] - droga przebyta po 9 s. [latex]s=frac{at_{10} ^2}{2}-frac{at_{9} ^2}{2}\ s=frac{5frac{m}{s^2}cdot (10s) ^2}{2}-frac{5frac{m}{s^2}cdot (9s) ^2}{2}\ s=frac{5cdot 100m}{2}-frac{5cdot 81m}{2}\ s=250m-202,5m\ s=47,5m[/latex] [latex]v_{sr}=frac{s_{10}}{t_{10}}\ v_{sr}=frac{250m}{10s}\ v_{sr}=25 frac{m}{s}[/latex] Odp. [latex]s=47,5m[/latex] [latex]v_{sr}=25frac{m}{s}[/latex] Pozdrawiam :) Jeśli pomogłem, proszę o naj. Pixelmaster
