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Question-163993




Question Number 163993 by mathls last updated on 12/Jan/22
Commented by mathls last updated on 12/Jan/22
please help
$${please}\:{help} \\ $$
Commented by mr W last updated on 12/Jan/22
relative acceleration =0  relative speed = 25 m/s  relative position at beginning =100 m  25t=100 ⇒t=4 s  i.e. after 4 seconds the distance  between both is zero.
$${relative}\:{acceleration}\:=\mathrm{0} \\ $$$${relative}\:{speed}\:=\:\mathrm{25}\:{m}/{s} \\ $$$${relative}\:{position}\:{at}\:{beginning}\:=\mathrm{100}\:{m} \\ $$$$\mathrm{25}{t}=\mathrm{100}\:\Rightarrow{t}=\mathrm{4}\:{s} \\ $$$${i}.{e}.\:{after}\:\mathrm{4}\:{seconds}\:{the}\:{distance} \\ $$$${between}\:{both}\:{is}\:{zero}. \\ $$
Commented by mathls last updated on 13/Jan/22
what is the practice?
$${what}\:{is}\:{the}\:{practice}? \\ $$
Commented by mr W last updated on 13/Jan/22
my practice is what i wrote.   relatively both objects move with a  relative speed 25m/s towards each  other. since they have a distance of  100 m at beginning, therefore after  4 seconds, their distance is zero.
$${my}\:{practice}\:{is}\:{what}\:{i}\:{wrote}.\: \\ $$$${relatively}\:{both}\:{objects}\:{move}\:{with}\:{a} \\ $$$${relative}\:{speed}\:\mathrm{25}{m}/{s}\:{towards}\:{each} \\ $$$${other}.\:{since}\:{they}\:{have}\:{a}\:{distance}\:{of} \\ $$$$\mathrm{100}\:{m}\:{at}\:{beginning},\:{therefore}\:{after} \\ $$$$\mathrm{4}\:{seconds},\:{their}\:{distance}\:{is}\:{zero}. \\ $$
Commented by mr W last updated on 13/Jan/22
maybe you would do as following.  position of object 1:  y=25t−((gt^2 )/2)  position of object 2:  y=100−((gt^2 )/2)  when both objects pass side by side,  25t−((gt^2 )/2)=100−((gt^2 )/2)  ⇒25t=100  ⇒t=((100)/(25))=4 seconds
$${maybe}\:{you}\:{would}\:{do}\:{as}\:{following}. \\ $$$${position}\:{of}\:{object}\:\mathrm{1}: \\ $$$${y}=\mathrm{25}{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${position}\:{of}\:{object}\:\mathrm{2}: \\ $$$${y}=\mathrm{100}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${when}\:{both}\:{objects}\:{pass}\:{side}\:{by}\:{side}, \\ $$$$\mathrm{25}{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{100}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{25}{t}=\mathrm{100} \\ $$$$\Rightarrow{t}=\frac{\mathrm{100}}{\mathrm{25}}=\mathrm{4}\:{seconds} \\ $$

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