A particle moves along a straight line and its position at time t is given by s(t)=2t^3–18t^2+30t where s is measured in meters and t in seconds.
What is the TOTAL distance travelled by the particle between time 0 and 12?
2007-09-20 10:07:09 · 6 answers · asked by Bikki 3 in Science & Mathematics ➔ Mathematics
Total distance= s(12)-s(0). s(0)=0 and you can evaluate the cubic in t for t=12. (12^3=1728)
2007-09-20 10:12:06 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
at t=0 s = 0
at t =12 s = 2(12)^3 -18(12)^2 +30(12) = 1224 meters.
If the particle moves along a straight line, then 1224 meters is the answer. If the particle is moving along the curve 2t^3-18t^2-30t, then that is a different problem. But you said a straight line.
2007-09-20 10:22:55 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
The answer is 1352 m, Sahsjing got it correct .... good job Sahsjing ... when I eneterd it on my TI graphic, I pushed the abs value button, under Math menu, but I guess it didnt register, so I thought I had the absolute value, when in fact , I did not .... should have paid attention to what was on my screen :-(
ironduke.... even if it moves along a straight line, it can have a position function as given in the question ... the function given just represents the position of the object along the line at any given time, .... the important question is, does the object ever move backwards ?? ... and since it does, then we need to take this into account when finding the total distance, which can't be done by taking s(12) - s(0) .....
To solve the problem, take:
int ( abs(velocity) ) from 0 to 12 is 1,352 m, where velocity = 6t^2 - 36t + 30, as done by Sahsjing above ...
by taking s(12) - s(0), which equals 1,224 m, you miss the fact that at times the object moves ' backwards", ...
s(12) - s( 0 ) will find the 'net distance' not the total distance... to do it this way, by using position, you would need to add up all the + distances [ "displacements" ], and then all the ' negative' distances [ backwards ], and take the absolute value of these neg. distances, before adding them together... the object goes backwards from about 2.2 sec to 6.8 sec, and so that part of it's distance results in a negative value, which reduces the total answer to the incorrect 1224 m ....
2007-09-20 10:26:46 · answer #3 · answered by Mathguy 5 · 0⤊ 0⤋
s'(t) = 6t^2-36t+30 = 6(t-1)(t-5)
TOTAL distance
= ∫|s'(t)| dt, t from 0 to 12
= ∫6t^2-36t+30 dt, t from 0 to 12 - 2∫6t^2-36t+30 dt, t from 1 to 5
= 1352
--------------
mikey,
You need to double check your work. Remember the total distance is accumulated along the path, with each segment being positive.
ironduke8159,
It can be a straight line just as you run 100 m, your speed is not a constant.
2007-09-20 10:12:12 · answer #4 · answered by sahsjing 7 · 0⤊ 1⤋
tan = y/x => opposite element = y and adjacent element = x hypotenuse = ?(x^2 + y^2) sin = y/?(x^2 + y^2) and csc = ?(x^2 + y^2) / y cos = x /?(x^2 + y^2) and sec = ?(x^2 + y^2) /x cot = a million/tan = x /y
2016-10-05 02:10:14 · answer #5 · answered by ? 4 · 0⤊ 0⤋
hell..
sorry...i really did try to figure this out...
i wouldnt consider this Basic...must be after my time.
2007-09-20 10:14:01 · answer #6 · answered by leroy jenkins 2 · 0⤊ 0⤋