![Create a visually appealing science-themed image that represents the principle of particle acceleration. The scene should include a particle moving along its path, demonstrating the changes in velocity and position over time. Include elements such as motion arrows, distance markers, and time indicators to concretely represent the concept, noting the values of -8m/s2 for acceleration, x=20m for position at t=4s, and x=4m for position when velocity is 16m/s. Do not include any textual elements.](/images/questions/1239728/1024x1024.png)
The acceleration of a particle is defined by the relation a=-8m/s2. Knowing that x=20m when t=4 and that x=4m when v=16m/s. Determine. A)the time when the velocity is zero. B)the velocity and total distance traveled when t=11s.
v = Vi - 8 t
x = Xi + Vi t - 4 t^2
20 = Xi + Vi (4) - 4(16)
20 = Xi + 4 Vi - 64
Xi + 4 Vi = 84 so Vi = (84-Xi)/4
16 = Vi - 8 t
16 = (84-Xi)/4 - 8 t (eqn 1)
4 = Xi + [(84-Xi) /4] t - 4 t^2 (eqn 2)
from eqn 1
64 = 84-Xi - 32 t
so
Xi + 32 t = 20
Xi = (20 - 32 t)
substitute that in eqn 2
4 = 20-32t+[(84-20+32t)]t/4 -4 t^2
4 = 20 - 32 t +16 t +8t^2 -4t^2
0 = 4 t^2 - 16 t + 16
t^2 -4 t + 4 = 0
(t-2)(t-2) = 0
t = 2 seconds when v = 16 and x = 4
Xi = 20 -32(2) = 20 - 64 = -44 at t = 0
16 = Vi - 8 t = Vi -16
so Vi = 32
WHEW !
SO IN the END
v = 32 - 8 t
and
x = -44 +32 t - 4 t^2
so
v = 0 when t = 4
PLUG in for part 2
BUT BEWARE
v is easy at 11 seconds
BUT for distance, some may be negative
because v = 0 between t = 0 at t = 4
you must add ABSOLUTE VALUES of distance before and after t = 4
( IT ASKED FOR DISTANCE NOT DISPLACEMENT)
(trying to trick you)
a = -8
v = -8t+c1
x = -4t^2 + c1*t + c2
x(4)=20, so
-64 + 4c1 + c2 = 20
when v=16,
-8t+c1=16, or t = (c1-16)/8
at that time, x=4, so
-4((c1-16)/8)^2 + c1(c1-16)/8 + c2 = 4
Solve those two equations for c1 and c2, and you get
c1 = 32
c2 = -44
That means that
a = -8
v = -8t+32
x = -4t^2+32t-44
check: v=16 when t=2
x(2) = 4
x(4) = 20
I expect that now you can answer the questions...