Question Number 198555 by Tawa11 last updated on 21/Oct/23
Commented by ajfour last updated on 23/Oct/23
$${both}\:{answers}\:\:\:{t}=\frac{\mathrm{17}}{\mathrm{3}}\:{s}\:\:\:{or}\:\:{t}=\mathrm{5}{s} \\ $$$${i}\:{would}\:{consider}\:{right}. \\ $$
Commented by Tawa11 last updated on 24/Oct/23
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by Tawa11 last updated on 21/Oct/23
A car 4m long moving at a velocity of 25m/s was
beside a lorry 20m long with velocity 19m/s.
At t = 0, the distance between them was 10m.
How long will it take the car to overtake the lorry.
(a) 2s
(b) 9s
(c) 5s
(d) 3s
beside a lorry 20m long with velocity 19m/s.
At t = 0, the distance between them was 10m.
How long will it take the car to overtake the lorry.
(a) 2s
(b) 9s
(c) 5s
(d) 3s
Commented by mr W last updated on 22/Oct/23
$${answer}\:\left({c}\right) \\ $$
Commented by Tawa11 last updated on 23/Oct/23
$$\mathrm{relative}\:\mathrm{velocity}\:=\:\mathrm{25}\:−\:\mathrm{19}\:\:=\:\:\mathrm{6m}/\mathrm{s} \\ $$$$\mathrm{time}\:\mathrm{to}\:\mathrm{cover}\:\mathrm{10}\:\mathrm{m} \\ $$$$\:\mathrm{t}\:\:=\:\:\mathrm{10}/\mathrm{6}\:\:=\:\:\mathrm{1}.\mathrm{67s} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{20m} \\ $$$$\mathrm{relative}\:\mathrm{velocity}\:=\:\mathrm{25}\:−\:\mathrm{19}\:\:=\:\:\mathrm{6m}/\mathrm{s} \\ $$$$\:\mathrm{t}\:\:=\:\:\mathrm{20}/\mathrm{6}\:\:=\:\:\mathrm{3}.\mathrm{33s} \\ $$$$\mathrm{total}\:\mathrm{time} \\ $$$$\:=\:\:\:\mathrm{1}.\mathrm{67}\:\:+\:\:\mathrm{3}.\mathrm{33} \\ $$$$=\:\:\:\:\mathrm{5s} \\ $$
Commented by Tawa11 last updated on 23/Oct/23
$$\mathrm{Sir},\:\mathrm{is}\:\mathrm{this}\:\mathrm{correct}? \\ $$
Commented by mr W last updated on 23/Oct/23
$${actually}\:{this}\:{should}\:{be}\:{an}\:{easy}\: \\ $$$${question}\:{to}\:{you},\:{i}'{d}\:{like}\:{to}\:{know}\:{what} \\ $$$${is}\:{your}\:{own}\:{solution}. \\ $$
Commented by mr W last updated on 23/Oct/23
$${to}\:{be}\:{exact}, \\ $$$${overtake}\:{means}\:{changing}\:{from}\:{the}\: \\ $$$${state}\:“{behind}\:{something}''\:{to}\:{the} \\ $$$${state}\:“{in}\:{front}\:{of}\:{this}\:{something}''. \\ $$$${that}\:{is}\:{to}\:{say},\:{for}\:{the}\:{car}\:{it}\:{has} \\ $$$${finished}\:{the}\:“{overtaking}''\:{when}\:{the} \\ $$$${end}\:{of}\:{the}\:{car}\:{has}\:{reached}\:{the}\:{front}\: \\ $$$${of}\:{the}\:{lorry}.\: \\ $$$${that}\:{means}\:{you}\:{have}\:{not}\:{considered} \\ $$$${the}\:{length}\:{of}\:{the}\:{car}\:{in}\:{your}\:{solution}. \\ $$
Commented by Tawa11 last updated on 23/Oct/23
$$\mathrm{ohh} \\ $$
Commented by Tawa11 last updated on 23/Oct/23
$$\mathrm{Vt}\:−\:\mathrm{vt}\:\:=\:\:\mathrm{L}\:\:+\:\:\mathrm{d}\:\:+\:\:\mathrm{l} \\ $$$$\mathrm{t}\left(\mathrm{25}\:\:−\:\:\mathrm{19}\right)\:\:\:=\:\:\:\mathrm{20}\:\:+\:\:\mathrm{10}\:\:+\:\:\mathrm{4} \\ $$$$\:\:\:\mathrm{6t}\:\:\:=\:\:\:\mathrm{34} \\ $$$$\:\:\:\:\:\:\mathrm{t}\:\:\:=\:\:\:\frac{\mathrm{34}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\mathrm{t}\:\:\:=\:\:\:\mathrm{5}.\mathrm{7s} \\ $$
Commented by Tawa11 last updated on 23/Oct/23
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{check}. \\ $$
Commented by mr W last updated on 23/Oct/23
$${but}\:{this}\:{answer}\:{is}\:{not}\:{given}\:?! \\ $$
Commented by Tawa11 last updated on 23/Oct/23
$$\mathrm{Yes}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 23/Oct/23
$$\mathrm{That}\:\mathrm{is}\:\mathrm{why}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{the}\:\mathrm{question}\:\mathrm{here}. \\ $$
Commented by mr W last updated on 23/Oct/23
$${i}\:{have}\:{guessed}\:{this}.\:{otherwise}\:{you} \\ $$$${won}'{t}\:{post}\:{this}\:“{easy}''\:{question}. \\ $$
Commented by mr W last updated on 23/Oct/23
Commented by Tawa11 last updated on 23/Oct/23
$$\mathrm{Show}\:\mathrm{me}\:\mathrm{better}\:\mathrm{way}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 23/Oct/23
$$\mathrm{0}\leqslant{c}<\mathrm{10}{m} \\ $$$$\Delta{t}=\frac{\mathrm{4}+{c}+\mathrm{20}}{\mathrm{25}−\mathrm{19}}=\frac{\mathrm{24}+{c}}{\mathrm{6}}\geqslant\frac{\mathrm{24}+\mathrm{0}}{\mathrm{6}}=\mathrm{4}\:{s} \\ $$$$\Delta{t}=\frac{\mathrm{24}+{c}}{\mathrm{6}}<\frac{\mathrm{24}+\mathrm{10}}{\mathrm{6}}=\mathrm{5}.\mathrm{7}\:{s} \\ $$$$\mathrm{4}{s}\:\leqslant\Delta{t}<\:\mathrm{5}.\mathrm{7}{s} \\ $$$$\Rightarrow\left({c}\right)\:{is}\:{the}\:{only}\:{possible}\:{right}\:{answer}. \\ $$$${with}\:\Delta{t}=\mathrm{5}:\:\frac{\mathrm{24}+{c}}{\mathrm{6}}=\mathrm{5}\:\Rightarrow{c}=\mathrm{6}\:{m} \\ $$$${i}.{e}.\:{the}\:{distance}\:{between}\:{lorry}\:{and} \\ $$$${car}\:{was}\:\mathrm{10}\:{m}\:{and}\:{the}\:{car}\:{was}\:\mathrm{6}\:{m} \\ $$$${behind}\:{the}\:{lorry}\:{in}\:{driving}\:{direction}. \\ $$
Commented by Tawa11 last updated on 24/Oct/23
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$