Revised as per suggestions in the posts.
1) A Flooble force causes a constant acceleration of 0.1 meters/sec/sec until its speed is 20 meters/sec
v1^2 - v0^2 = 2 a x1
20^2 = 2 (0.1) x1
x1 = 2000 m
x1 = (1/2) a t1^2
2000 = (1/2) 0.1 t1^2
t1 = 200 s
x1 = 2000 m, t1 = 200 s, v1=20 m/s
2) There is a Flooble zone that lowers the speed limit. Therefore, the ball does a constant deceleration to one half of its speed over a distance of 750 meters
v2^2 = v1^2 + 2 a x
10^2 = 20^2 + 2 a 750
a = -0.2 m/s^2
v2 = v1 + a t
10 = 20 + (-0.2) (50)
dt = 50 s
dx = 750 m
x2 = 2750 m, t2 = 250 s, v2 = 10 m/s
3) A Floobleoid starts pushing the ball. This adds a 5 Newton assist to accelerate the ball for 320 seconds
a = F/m = 5/80 = 0.0625 m/s^2
dt = 320s
t3 = 250 + 320 = 570 s
v3 = v2 + a dt = 10 + 0.0625 (320) = 30 m/s
x3 = x2 + v2 dt + 1/2 a dt^2
= 2750 + 10 (320) + (1/2) (0.0625) 320^2
= 9150 m
x3 = 9150 m, t3 = 570 s, v3 = 30 m/s
4) The ball rolls off the edge of the Flooble Canyon, which is a sheer vertical cliff of height exactly 490.35 meter. After it hits the bottom in the flat and level Flooble Valley, it's vertical speed is immediately zero and it continues on for an additional 150 seconds
Are we here to assume that the airborne ball is no longer "rolling"?
While it will have an x and y displacement during its falling time tf of xf, yf, I will assume these do not contribute to its total "rolling" distance.....
(if I am wrong - then add xf, or worse, add the length of the ball's parabolic path) BTW - where ever did the y-velocity component of energy go?
yf = 490.35 = 1/2 g tf^2 = 1/2 9.807 tf^2
tf = sqrt [490.35/ (0.5 9.807)] = 10 s
xf = v3 dt = 30 (10) = 300 m
after hitting it rolls on for dt = 150 s.
v4 = v3
dt = 150 s
dx = v3 dt = 30 (150) = 4500 m
Without the "falling x" of 300 m ....
x4 = x3 + dx = 9150 + 4500 = 13650 m
t4 = t3 + tf + dt = 570 + 10 + 150 = 730 s
x4 = 13650 m, t4 = 730 s, v4 = 30 m/s
5) The ball encounters a Flooble battery and immediately gains 30 Watt-hours of kinetic energy. It then rolls on for 1200 meters
30 W hours = 30 W (3600 s) = 30 (3600) J = 108000 J
Currently the K.E. of the ball is (1/2) m v4^2 =
(1/2) 80 (30)^2 = 36000 J
The new total K.E. is E5 = 36000 J + 108000 J =
= 144000 J =1/2 m v5^2, so v5=60 m/s
dx = dt v5
dt = dx/v5 = 1200 / 60 = 20 s
t5 = t4 + dt = 730 + 20 = 750 s
x5 = x4 + dx = 13650 + 1200 = 14850 m
x5 = 14850 m, t5 = 750 s, v5 = 60 m/s
6) The ball encounters a null-Flooble-energy-field and experiences no forces of any kind for 60 seconds
v6 = v5
t6 = t5 + 60 s = 750 + 60 = 810 s
x6 = x5 + v5 dt = 14850 m + 60 (60) = 18450 m
x6 = 18450 m, t6 = 810 s, v6 = 60 m/s
7) The ball Flooble-decelerates at a constant rate to a stop in 15 seconds, or, maybe in 500 meters, whichever stops the ball more quickly.
I took this to mean choose from two possible different deceleration rates: A rate which stops the ball in 15 seconds is a15. A rate which stops it in 500 meters is a500. We look at each.
v7^2 - v6^2 = 2 a500 dx
0 - 60^2 = 2 a500 (500)
a500 = 60^2 / 1000 = -3.6 m/s^2
v7 = v6 + a500 dt
0 = 60 - 3.6 dt
dt = 60 / 3.6 = 16.666 s
So, the a15 deceleration scheme is quicker.
v7 = 60 - a15 t7
0 = 60 - a15 15
a15 = -4 m/s^2
v7^2 - v6^2 = 2 a15 dx
v6^2 = 2 (4) dx
dx = 60^2/8 =450
x7 = x6 + dx = 18450 + 450 = 18900 m
t7 = t6 + dt = 810 + 15 = 825 s
x7 = 18900 m, t7 = 825 m/s, v7 = 0
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