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Mathematics 8 Online
OpenStudy (anonymous):

MEDAL WILL BE GIVEN how do i work out an estimate for the instantaneous velocity when given one number?

OpenStudy (anonymous):

one number for what? position?

OpenStudy (anonymous):

@sky202 don't just dump a problem on us... we need your input.

OpenStudy (anonymous):

okay well i just need an answer or steps in general but since you insist i was given....1 Let s(t) = 6t − 2t3 be the displacement function of a particle moving in a straight line, where t is in seconds and s is in metres. a) Find the average velocity for the time interval [0, 1] b) . Find the average velocity for the time interval [0.8, 1]. i have worked out a to be 4 meters per second

OpenStudy (anonymous):

@Peter14 come back :(

OpenStudy (anonymous):

@sky202. You can calculate the total displacement in the timeframe given by integrating s(t) over the time interval. This gives you the total distance D and the span of the time interval provides you with the travel time T it took the particle to cover this distance. It's average velocity follows from total distance / travel time, being D/T. The same mechanism applies to your b) part of the question, but this time with different values for the time interval. I leave the actual calculations for you to solve, but don't hesitate to ask if you need further support.

OpenStudy (anonymous):

i have the displacement i want to know how to work out part b in simpler terms please

OpenStudy (anonymous):

What did you find after integrating s(t) over t ? Must be something like S(t) = 3*t^2 - 0.5*t^4, true ?

OpenStudy (anonymous):

So applying S(t) (the integral of s(t)) to the time interval of [t=0:t=1] gives the following: S(t=1) - S(t=0) = 2.5 - 0 = 2.5, which is the average speed over [0:1]. Same calculation for [t=0.8 : t=1], so S(t=1) - S(t=0.8)...OK ?

OpenStudy (fifciol):

average velocity is very simple: \[\Delta V=\frac{ \Delta s }{ \Delta t }=\frac{ s(1)-s(0,8) }{ 1-0,8 }\]

OpenStudy (anonymous):

how do i attach a photo ?

OpenStudy (anonymous):

ill try thats gimme 1 sec

OpenStudy (fifciol):

your answer 4m/s is correct in a

OpenStudy (anonymous):

i know but i got for part b 1.12 like i got originally but my book says im wrong and that it is -1/2 maybe its wrong i have no idea? could you possibly try it and see what you get :)

OpenStudy (fifciol):

i find 1,12

OpenStudy (anonymous):

yaaaaaaaay so we are right :D awesome thanks very much heres your medal :)

OpenStudy (fifciol):

yw:)

OpenStudy (anonymous):

sorry, I would have gotten it, but OS wasn't working for me. NLCircle and Fifciol did a good job, possibly better than I could have.

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