Edits to “Solution”, “Answer”
en/1.2.21.md
+5 −5
| ### Statement | |||
| $1.2.21^*.$ The scheduled departure time of the train is $12.00$. It's $12.00$ on your watch, but the penultimate car that moves past you during time $t_1$ is already starting to pass by you. The last car passes you during $t_2$. The train left on time and is moving equidistant. How far behind is your watch? | |||
| ### Solution | |||
| $$ | |||
| l=\nu_{1}t_{1} +\frac{at_{1}^{2}}{2} | |||
| @@ -9,25 +9,25 @@Solution | |||
| $$ | |||
| $$ | |||
| − | l=(\nu_{1}+at_{1})t_{2} +\frac{at_{ | ||
| + | l=(\nu_{1}+at_{1})t_{2} +\frac{at_{2}^{2}}{2} | ||
| $$ | |||
| $$ | |||
| − | (\nu_{1}+at_{1})t_{2} +\frac{at_{ | ||
| + | (\nu_{1}+at_{1})t_{2} +\frac{at_{2}^{2}}{2} = \nu_{1}t_{1} +\frac{at_{1}^{2}}{2} | ||
| $$ | |||
| Given that $\nu_{1} = a{d}t$: | |||
| $$ | |||
| − | a{d}t(t_{1}-t_{2}) =\frac{a(t_{2} | ||
| + | a{d}t(t_{1}-t_{2}) =\frac{a(t_{2}^{2}+2t_{1}t_{2}-t_{1}^{2})}{2} | ||
| $$ | |||
| $$ | |||
| − | {d}t=\fbox{$\frac{t_{2} | ||
| + | {d}t=\fbox{$\frac{t_{2}^{2}+2t_{1}t_{2}-t_{1}^{2}}{2(t_{1}-t_{2})}$} | ||
| $$ | |||
| #### Answer | |||
| $$ | |||
| − | \frac{t_{2} | ||
| + | \frac{t_{2}^{2}+2t_{1}t_{2}-t_{1}^{2}}{2(t_{1}-t_{2})} | ||
| $$ | |||
| ### Statement | ### Statement | ||
| $1.2.21^*.$ The scheduled departure time of the train is $12.00$. It's $12.00$ on your watch, but the penultimate car that moves past you during time $t_1$ is already starting to pass by you. The last car passes you during $t_2$. The train left on time and is moving equidistant. How far behind is your watch? | $1.2.21^*.$ The scheduled departure time of the train is $12.00$. It's $12.00$ on your watch, but the penultimate car that moves past you during time $t_1$ is already starting to pass by you. The last car passes you during $t_2$. The train left on time and is moving equidistant. How far behind is your watch? | ||
| ### Solution | ### Solution | ||
| $$ | $$ | ||
| l=\nu_{1}t_{1} +\frac{at_{1}^{2}}{2} | l=\nu_{1}t_{1} +\frac{at_{1}^{2}}{2} | ||
| @@ -9,25 +9,25 @@Solution | |||
| $$ | $$ | ||
| $$ | $$ | ||
| l=(\nu_{1}+at_{1})t_{2} +\frac{at_{ |
l=(\nu_{1}+at_{1})t_{2} +\frac{at_{2}^{2}}{2} | ||
| $$ | $$ | ||
| $$ | $$ | ||
| (\nu_{1}+at_{1})t_{2} +\frac{at_{ |
(\nu_{1}+at_{1})t_{2} +\frac{at_{2}^{2}}{2} = \nu_{1}t_{1} +\frac{at_{1}^{2}}{2} | ||
| $$ | $$ | ||
| Given that $\nu_{1} = a{d}t$: | Given that $\nu_{1} = a{d}t$: | ||
| $$ | $$ | ||
| a{d}t(t_{1}-t_{2}) =\frac{a(t_{2} |
a{d}t(t_{1}-t_{2}) =\frac{a(t_{2}^{2}+2t_{1}t_{2}-t_{1}^{2})}{2} | ||
| $$ | $$ | ||
| $$ | $$ | ||
| {d}t=\fbox{$\frac{t_{2} |
{d}t=\fbox{$\frac{t_{2}^{2}+2t_{1}t_{2}-t_{1}^{2}}{2(t_{1}-t_{2})}$} | ||
| $$ | $$ | ||
| #### Answer | #### Answer | ||
| $$ | $$ | ||
| \frac{t_{2} |
\frac{t_{2}^{2}+2t_{1}t_{2}-t_{1}^{2}}{2(t_{1}-t_{2})} | ||
| $$ | $$ | ||