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Question Number 171614 by Tawa11 last updated on 18/Jun/22

A particle is moving along a straight line such that it's position from a fixed point is S = ( 12 - 15t² + 5t³ )m where t is in seconds. Determine: A. Total distance travelled by the particle from t = 1sec to t = 3sec B. The average speed of the particle during this time.\\n

Commented byTawa11 last updated on 18/Jun/22

Is total distance  14m?

$$\mathrm{Is}\:\mathrm{total}\:\mathrm{distance}\:\:\mathrm{14m}? \\ $$

Commented byBeginner last updated on 18/Jun/22

total distance=s1+s2. substitute the parameters directly

$${total}\:{distance}={s}\mathrm{1}+{s}\mathrm{2}.\:{substitute}\:{the}\:{parameters}\:{directly} \\ $$ $$ \\ $$

Commented bymr W last updated on 18/Jun/22

30 m

$$\mathrm{30}\:{m} \\ $$

Commented bymr W last updated on 19/Jun/22

at 1 o′clock you are 2 km easterly   apart from your house,  at 2 o′clock you are 8 km westerly   apart from your house,  that means from 1 o′clock to 2 o′clock  you walked 10 km in direction  from east to west.    at 3 o′clock you are again 12 km easterly   apart from your house,  that means from 2 o′clock to 3 o′clock  you walked 20 km in direction  from west to east.    that means from 1 o′clock to 3 o′clock  you walked totally 30 km.

$${at}\:\mathrm{1}\:{o}'{clock}\:{you}\:{are}\:\mathrm{2}\:{km}\:{easterly}\: \\ $$ $${apart}\:{from}\:{your}\:{house}, \\ $$ $${at}\:\mathrm{2}\:{o}'{clock}\:{you}\:{are}\:\mathrm{8}\:{km}\:{westerly}\: \\ $$ $${apart}\:{from}\:{your}\:{house}, \\ $$ $${that}\:{means}\:{from}\:\mathrm{1}\:{o}'{clock}\:{to}\:\mathrm{2}\:{o}'{clock} \\ $$ $${you}\:{walked}\:\mathrm{10}\:{km}\:{in}\:{direction} \\ $$ $${from}\:{east}\:{to}\:{west}. \\ $$ $$ \\ $$ $${at}\:\mathrm{3}\:{o}'{clock}\:{you}\:{are}\:{again}\:\mathrm{12}\:{km}\:{easterly}\: \\ $$ $${apart}\:{from}\:{your}\:{house}, \\ $$ $${that}\:{means}\:{from}\:\mathrm{2}\:{o}'{clock}\:{to}\:\mathrm{3}\:{o}'{clock} \\ $$ $${you}\:{walked}\:\mathrm{20}\:{km}\:{in}\:{direction} \\ $$ $${from}\:{west}\:{to}\:{east}. \\ $$ $$ \\ $$ $${that}\:{means}\:{from}\:\mathrm{1}\:{o}'{clock}\:{to}\:\mathrm{3}\:{o}'{clock} \\ $$ $${you}\:{walked}\:{totally}\:\mathrm{30}\:{km}. \\ $$

Commented bymr W last updated on 19/Jun/22

you should know that the particle  doesn′t move in one direction only.  just study s=12−15t^2 +5t^3  in detail!

$${you}\:{should}\:{know}\:{that}\:{the}\:{particle} \\ $$ $${doesn}'{t}\:{move}\:{in}\:{one}\:{direction}\:{only}. \\ $$ $${just}\:{study}\:{s}=\mathrm{12}−\mathrm{15}{t}^{\mathrm{2}} +\mathrm{5}{t}^{\mathrm{3}} \:{in}\:{detail}! \\ $$

Commented bymr W last updated on 19/Jun/22

Commented byTawa11 last updated on 19/Jun/22

God bless you sirs

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sirs} \\ $$

Commented bymr W last updated on 19/Jun/22

Distance traveled is the total length   of the path traveled between two  positions.  In current example the particle traveled  a path of 30m length in the second  and third second. so the distance  traveled is 30m.

$$\mathrm{\color{mathred}{D}\color{mathred}{i}\color{mathred}{s}\color{mathred}{t}\color{mathred}{a}\color{mathred}{n}\color{mathred}{c}\color{mathred}{e}}\color{mathred}{\:}\mathrm{\color{mathred}{t}\color{mathred}{r}\color{mathred}{a}\color{mathred}{v}\color{mathred}{e}\color{mathred}{l}\color{mathred}{e}\color{mathred}{d}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{total}\:\mathrm{length}\: \\ $$ $$\mathrm{of}\:\mathrm{the}\:\mathrm{path}\:\mathrm{traveled}\:\mathrm{between}\:\mathrm{two} \\ $$ $$\mathrm{position}{s}. \\ $$ $${In}\:{current}\:{example}\:{the}\:{particle}\:{traveled} \\ $$ $${a}\:{path}\:{of}\:\mathrm{30}{m}\:{length}\:{in}\:{the}\:{second} \\ $$ $${and}\:{third}\:{second}.\:{so}\:{the}\:{distance} \\ $$ $${traveled}\:{is}\:\mathrm{30}{m}. \\ $$

Commented bymr W last updated on 19/Jun/22

Commented bymr W last updated on 19/Jun/22

an object moved from A to B   following the red path with length s.  displacement=AB^(→)   distance=∣AB^(→) ∣=d  distance traveled=length of path=s

$${an}\:{object}\:{moved}\:{from}\:{A}\:{to}\:{B}\: \\ $$ $${following}\:{the}\:{red}\:{path}\:{with}\:{length}\:{s}. \\ $$ $${displacement}=\overset{\rightarrow} {{AB}} \\ $$ $${distance}=\mid\overset{\rightarrow} {{AB}}\mid={d} \\ $$ $${distance}\:{traveled}={length}\:{of}\:{path}={s} \\ $$

Commented byTawa11 last updated on 21/Jun/22

Wow, I just finished studying your concept sir.  Here displacement still end up in  10m  That is     (−   8   −   2)  +  [12   −  (−  8)]   =   10m.  And  Distance   =   8  +  2  +  20   =   30m

$$\mathrm{Wow},\:\mathrm{I}\:\mathrm{just}\:\mathrm{finished}\:\mathrm{studying}\:\mathrm{your}\:\mathrm{concept}\:\mathrm{sir}. \\ $$ $$\mathrm{Here}\:\mathrm{displacement}\:\mathrm{still}\:\mathrm{end}\:\mathrm{up}\:\mathrm{in}\:\:\mathrm{10m} \\ $$ $$\mathrm{That}\:\mathrm{is}\:\:\:\:\:\left(−\:\:\:\mathrm{8}\:\:\:−\:\:\:\mathrm{2}\right)\:\:+\:\:\left[\mathrm{12}\:\:\:−\:\:\left(−\:\:\mathrm{8}\right)\right]\:\:\:=\:\:\:\mathrm{10m}. \\ $$ $$\mathrm{And} \\ $$ $$\mathrm{Distance}\:\:\:=\:\:\:\mathrm{8}\:\:+\:\:\mathrm{2}\:\:+\:\:\mathrm{20}\:\:\:=\:\:\:\mathrm{30m} \\ $$

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