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.

Answer 1

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)

Answer 2

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...

Answer 3

we agree !

Answer 4

I started reading your solution with trepidation. When I saw that we came up with the same solution, i breathed a huge sigh of relief!

Answer 5

me too :)

Answer 6

Thanks guys, I really appreciate your effort.

Answer 7

You are welcome. Actually that was fun :)