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Question-198555

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Question Number 198555 by Tawa11 last updated on 21/Oct/23
Commented by ajfour last updated on 23/Oct/23
both answers t=((17)/3) s or t=5s i would consider right.
$${both}\:{answers}\:\:\:{t}=\frac{\mathrm{17}}{\mathrm{3}}\:{s}\:\:\:{or}\:\:{t}=\mathrm{5}{s} \\ $$$${i}\:{would}\:{consider}\:{right}. \\ $$
Commented by Tawa11 last updated on 24/Oct/23
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
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
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

Commented by mr W last updated on 22/Oct/23
answer (c)
$${answer}\:\left({c}\right) \\ $$
Commented by Tawa11 last updated on 23/Oct/23
relative velocity = 25 − 19 = 6m/s time to cover 10 m t = 10/6 = 1.67s for 20m relative velocity = 25 − 19 = 6m/s t = 20/6 = 3.33s total time = 1.67 + 3.33 = 5s
$$\mathrm{relative}\:\mathrm{velocity}\:=\:\mathrm{25}\:−\:\mathrm{19}\:\:=\:\:\mathrm{6m}/\mathrm{s} \\ $$$$\mathrm{time}\:\mathrm{to}\:\mathrm{cover}\:\mathrm{10}\:\mathrm{m} \\ $$$$\:\mathrm{t}\:\:=\:\:\mathrm{10}/\mathrm{6}\:\:=\:\:\mathrm{1}.\mathrm{67s} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{20m} \\ $$$$\mathrm{relative}\:\mathrm{velocity}\:=\:\mathrm{25}\:−\:\mathrm{19}\:\:=\:\:\mathrm{6m}/\mathrm{s} \\ $$$$\:\mathrm{t}\:\:=\:\:\mathrm{20}/\mathrm{6}\:\:=\:\:\mathrm{3}.\mathrm{33s} \\ $$$$\mathrm{total}\:\mathrm{time} \\ $$$$\:=\:\:\:\mathrm{1}.\mathrm{67}\:\:+\:\:\mathrm{3}.\mathrm{33} \\ $$$$=\:\:\:\:\mathrm{5s} \\ $$
Commented by Tawa11 last updated on 23/Oct/23
Sir, is this correct?
$$\mathrm{Sir},\:\mathrm{is}\:\mathrm{this}\:\mathrm{correct}? \\ $$
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.
$${actually}\:{this}\:{should}\:{be}\:{an}\:{easy}\: \\ $$$${question}\:{to}\:{you},\:{i}'{d}\:{like}\:{to}\:{know}\:{what} \\ $$$${is}\:{your}\:{own}\:{solution}. \\ $$
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.
$${to}\:{be}\:{exact}, \\ $$$${overtake}\:{means}\:{changing}\:{from}\:{the}\: \\ $$$${state}\:“{behind}\:{something}''\:{to}\:{the} \\ $$$${state}\:“{in}\:{front}\:{of}\:{this}\:{something}''. \\ $$$${that}\:{is}\:{to}\:{say},\:{for}\:{the}\:{car}\:{it}\:{has} \\ $$$${finished}\:{the}\:“{overtaking}''\:{when}\:{the} \\ $$$${end}\:{of}\:{the}\:{car}\:{has}\:{reached}\:{the}\:{front}\: \\ $$$${of}\:{the}\:{lorry}.\: \\ $$$${that}\:{means}\:{you}\:{have}\:{not}\:{considered} \\ $$$${the}\:{length}\:{of}\:{the}\:{car}\:{in}\:{your}\:{solution}. \\ $$
Commented by Tawa11 last updated on 23/Oct/23
ohh
$$\mathrm{ohh} \\ $$
Commented by Tawa11 last updated on 23/Oct/23
Vt − vt = L + d + l t(25 − 19) = 20 + 10 + 4 6t = 34 t = ((34)/6) t = 5.7s
$$\mathrm{Vt}\:−\:\mathrm{vt}\:\:=\:\:\mathrm{L}\:\:+\:\:\mathrm{d}\:\:+\:\:\mathrm{l} \\ $$$$\mathrm{t}\left(\mathrm{25}\:\:−\:\:\mathrm{19}\right)\:\:\:=\:\:\:\mathrm{20}\:\:+\:\:\mathrm{10}\:\:+\:\:\mathrm{4} \\ $$$$\:\:\:\mathrm{6t}\:\:\:=\:\:\:\mathrm{34} \\ $$$$\:\:\:\:\:\:\mathrm{t}\:\:\:=\:\:\:\frac{\mathrm{34}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\mathrm{t}\:\:\:=\:\:\:\mathrm{5}.\mathrm{7s} \\ $$
Commented by Tawa11 last updated on 23/Oct/23
Sir, please check.
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{check}. \\ $$
Commented by mr W last updated on 23/Oct/23
but this answer is not given ?!
$${but}\:{this}\:{answer}\:{is}\:{not}\:{given}\:?! \\ $$
Commented by Tawa11 last updated on 23/Oct/23
Yes sir.
$$\mathrm{Yes}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 23/Oct/23
That is why I posted the question here.
$$\mathrm{That}\:\mathrm{is}\:\mathrm{why}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{the}\:\mathrm{question}\:\mathrm{here}. \\ $$
Commented by mr W last updated on 23/Oct/23
i have guessed this. otherwise you won′t post this “easy” question.
$${i}\:{have}\:{guessed}\:{this}.\:{otherwise}\:{you} \\ $$$${won}'{t}\:{post}\:{this}\:“{easy}''\:{question}. \\ $$
Commented by mr W last updated on 23/Oct/23
Commented by Tawa11 last updated on 23/Oct/23
Show me better way sir.
$$\mathrm{Show}\:\mathrm{me}\:\mathrm{better}\:\mathrm{way}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 23/Oct/23
0≤c<10m Δt=((4+c+20)/(25−19))=((24+c)/6)≥((24+0)/6)=4 s Δt=((24+c)/6)<((24+10)/6)=5.7 s 4s ≤Δt< 5.7s ⇒(c) is the only possible right answer. with Δt=5: ((24+c)/6)=5 ⇒c=6 m i.e. the distance between lorry and car was 10 m and the car was 6 m behind the lorry in driving direction.
$$\mathrm{0}\leqslant{c}<\mathrm{10}{m} \\ $$$$\Delta{t}=\frac{\mathrm{4}+{c}+\mathrm{20}}{\mathrm{25}−\mathrm{19}}=\frac{\mathrm{24}+{c}}{\mathrm{6}}\geqslant\frac{\mathrm{24}+\mathrm{0}}{\mathrm{6}}=\mathrm{4}\:{s} \\ $$$$\Delta{t}=\frac{\mathrm{24}+{c}}{\mathrm{6}}<\frac{\mathrm{24}+\mathrm{10}}{\mathrm{6}}=\mathrm{5}.\mathrm{7}\:{s} \\ $$$$\mathrm{4}{s}\:\leqslant\Delta{t}<\:\mathrm{5}.\mathrm{7}{s} \\ $$$$\Rightarrow\left({c}\right)\:{is}\:{the}\:{only}\:{possible}\:{right}\:{answer}. \\ $$$${with}\:\Delta{t}=\mathrm{5}:\:\frac{\mathrm{24}+{c}}{\mathrm{6}}=\mathrm{5}\:\Rightarrow{c}=\mathrm{6}\:{m} \\ $$$${i}.{e}.\:{the}\:{distance}\:{between}\:{lorry}\:{and} \\ $$$${car}\:{was}\:\mathrm{10}\:{m}\:{and}\:{the}\:{car}\:{was}\:\mathrm{6}\:{m} \\ $$$${behind}\:{the}\:{lorry}\:{in}\:{driving}\:{direction}. \\ $$
Commented by Tawa11 last updated on 24/Oct/23
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$

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