SOMEONE PLEASE HELP ME WITH THIS. A particle P moves in a straight line so that its displacement, S metres, from a fixed point O is given by S=4+15t-t³ where t is the time in seconds after passing through a point X on the line.
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Calculate the distance moved by the particle during the third second of its motion.

Answer is 4.72m. Can any genius show me the steps?

Question

SOMEONE PLEASE HELP ME WITH THIS. A particle P moves in a straight line so that its displacement, S metres, from a fixed point O is given by S=4+15t-t³ where t is the time in seconds after passing through a point X on the line.
--------------------------------------…
Calculate the distance moved by the particle during the third second of its motion.

Answer is 4.72m. Can any genius show me the steps?

ganeshie8 · Answer

familiar wid some calculus ?

anonymous · Answer

yes.

anonymous · Answer

not up to differential equations but yes i am familiar.

ganeshie8 · Answer

Good :) me too same. we can work it quickly just by using differentiation

ganeshie8 · Answer

first observe that the function has a zero slope between (2, 3)
lets find this exact point, where the slope is zero

ganeshie8 · Answer

$ S=4+15t-t^3$
$S' =  15 - 3t^2$

setting it to 0 and solving gives us $t  = \pm \sqrt{5}$

ganeshie8 · Answer

$ \sqrt{5}$ is the time   we're interested in

anonymous · Answer

beacuse at sqrt5 the particle is instantaneously at rest?

ganeshie8 · Answer

yup ! thats where, it changes its direction ! 
so for distance calculation, we need to calculate UP distance and DOWN distance both, between [2, 3]

ganeshie8 · Answer

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