Inactive Tutor answered • 08/10/23
v= t^2 -7t +10 = velocity at time t, v measured in meters per second, t measured in seconds
find
"distance traveled"
does that mean total distance traveled or displacement?
the particle goes forward from time t=-1 to t=2, for 3 seconds
then it goes backwards from t=2 to t=5 for 3 seconds
then it goes forward again from t=5 to t=7 for 2 seconds
to find 2 and 5, the U-turn points set v=0
t^2 -7t +10 =0
factor
(t-5)(t-2) = 0
set each factor equal to zero
t-5=0, t=5 seconds
t-2=0, t =2 seconds
t = 2 and 5 are when the particle changes direction
for displacement just add distance from t=-1 to 2 and t=5 to 7 but subtract distance during t=2 to t=5
or just find the distance from s(-1) to s(7)
for total distance traveled, sum all three distances, s(7)-s(5) + s(5)-s(2) + s(2)-s(1)
v(t) = s'(t)
s(t) = the integral of v(t) =
s = t^3/3 -7t^2/2 +10t
evaluate from -1 to 7
displacement = s(7)-s(-1)
=7^3/3 -7^3/2 +70 - ((-1)^3/3 -7(-1)^2 +10(-1))
=-343(2)/6-343(3)/6+420/6 +1/3+7/2+10
= 77/6 +83/6 = 13 5/6 + 12 5/6 = 26 2/3
= 160/6
= 26 2/3 meters for 8 seconds,= displacement (not total distance traveled), all at an average velocity of 3.4 m/s, that's averaging positive and negative speeds, so it's far less than the average absolute values of speed, back and forth
displacement is total distance traveled minus the backward motion's distance
backward distance = s(5)-s(2) = 4.5=27/6
77/6 + 83/6+27/6 -27/6
= 26 2/3 meters= displacement
for total distance traveled add twice the backward distance to the displacement
26 2/3 + 2(4 1/2)= 26 2/3 + 9
= 35 2/3 meters = total distance
total distance traveled = = s(7)-s(5) +2( s(5)-s(2)) + s(2)-s(-1)
the sum of 2 forward motions plus 1 backwards motion distance
s(5) = 5^3/3 -7(5^2)/2 +10(5) = 125/3 -175/2 + 50 = 50+250/6 -525/6 = 50-275/6= 25/6 = 4 1/6 m
s(2) = 2^3/3 -7(2^2)/2 +10(2) = 8/3-28/2 +20 = 8/3+6 = 8 2/3 m
s(5)-s(2) = 8 2/3 - 4 1/6= 4 1/2 multiply times 2 = 9
s(7)-s(5) = 77/6 -25/6 = 52/6 = 8 2/3
s(2)-s(-1) = 52/6 - -83/6 = 135/6 = 8 2/3 + 13 5/6 = 22 1/2
s(2)-s(-1) + s(7)-s(5) + s(5)-s(2) = 22 1//2 + 8 2/3 + 4 1/2
= 35 2/3 meters = total distance traveled
work the problem different ways and when they come out the same, odds are good the answers are correct
if different answers, at least one answer is wrong.
min speed of the particle is when v'=0
v' = 2t-7
t = 3.5
v(3.5) = (3.5)^2 -7(3.5) +10
= 12.25 +10 - 24.5 = - 1.25 m/s
max v = (-1)^2 -7(-1) +10
= 1+7+10 = 18 m/s which is faster than any track Olympic gold medalist
and it goes faster approaching infinity as t gets ever larger beyond the domain in this problem
v(8) =64-56+10 = 18 m/sec
v(8+) = 18+ m/sec
even exceeding the speed of light, causing time to slow down or go backwards? which makes velocity calculations suspect, difficult, impossible or very misleading. You may need to consult an Einsetin on this problem or actually a quantum physics expert would be better. . It's not realistic. Generally, an appeal to what's real is a check on the answer, but not in this problem. the use of negative time with t<0 is also a tip off there's something very strange about this problem. time doesn't go backwards.
It may help to graph v and s
v is an upward opening parabola, min=vertex = (3.5, about 6), x intercepts 2 and 5, y intercept =10
s is a cubic with two turning points, with relative max = (2, 8 2/3), min = (5,4 1/6) x-intercept= y intercept = the origin = (0,0)
other points (-1,-13 5/6), (7,12 5/6), (3.5, 26 2/3), left end behavior approaches negative infinity, right end behavior approaches positive infinity
