A-particle-is-moving-along-a-straight-line-such-that-it-s-position-from-a-fixed-point-is-S-12-15t-5t-m-where-t-is-in-seconds-Determine-A-Total-distance-travelled-by-the-particle-from-t-

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.

Commented by Tawa11 last updated on 18/Jun/22

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

Commented by Beginner last updated on 18/Jun/22

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

Commented by mr W last updated on 18/Jun/22

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

Commented by mr W last updated on 19/Jun/22

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

Commented by mr W last updated on 19/Jun/22

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

Commented by mr W last updated on 19/Jun/22

Commented by Tawa11 last updated on 19/Jun/22

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

Commented by mr W last updated on 19/Jun/22

$$\mathrm{Distance}\:\mathrm{traveled}\:\mathrm{is}\:\mathrm{the}\:\mathrm{total}\:\mathrm{length}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{path}\:\mathrm{traveled}\:\mathrm{between}\:\mathrm{two} \\ $$$$\mathrm{position}{s}. \\ $$$${In}\:{current}\:{example}\:{the}\:{particle}\:{traveled} \\ $$$${a}\:{path}\:{of}\:\mathrm{30}{m}\:{length}\:{in}\:{the}\:{second} \\ $$$${and}\:{third}\:{second}.\:{so}\:{the}\:{distance} \\ $$$${traveled}\:{is}\:\mathrm{30}{m}. \\ $$

Commented by mr W last updated on 19/Jun/22

Commented by mr W last updated on 19/Jun/22

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

Commented by Tawa11 last updated on 21/Jun/22

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

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