Posted by Madeline on .
Two particles move along the x -axis. For 0 is less than or equal to t is less than or equal to 6, the position of particle P at time t is given by p(t)=2cos((pi/4)t), while the position of particle R at time t is given by r(t)=t^3 -6t^2 +9t+3.
1. For 0 is less than or equal to t is less than or equal to 6, find all times t during which particle R is moving to the left. 2. for 0 is less than or equal to t is less than or equal to 6, find all times t during which the two particles travel in opposite directions. 3. Find the acceleration of particles P at time t=3. Is particle P speeding up, slowing down, or doing neither at time t=3? Explain your reasoning. 4. Show that during the interval (1,3), there must be at least one instant when the particle R must have a velocity of -2.
1. p '(t) = -2(π/4) sin(π/4t) , which is the velocity
to move to the left, p '(t) < 0 or negative, so
sin (π/4t) must be positive.
the period of sin(π/4t) is 2π/(π/4) = 8
so for the first 4 seconds the curve is above the x-axis or it is positive
so for 0 ≤ t ≤ 6, the particle is moving to the left for the first 4 seconds, then to the right for the next 2 seconds.
ARGGGG , it asked for particle R and I found it for P, Oh well, maybe we can use the above later on
r ' (t) = 3t^2 - 12t + 9
zeros are t = 1 and t = 3
so r'(t) is positive for t < 1 or t > 3 and is negative for 1 < t < 3
make a sketch of both derivatives on the same grid
look at when both curves are on the same side of the x-axis ---> going same direction
when both curves are on opposite sides of the x-axis --> same direction
From 0 to 1, R is moving to the right,
from 1 to 3, R is moving to the left
from 3 to 6 , R is moving to the right again.
So from 0 to 1, they are moving in the opposite directions
From 1 to 3, they are moving in the same direction , (both v's are negative
from 3 to 4 , they are moving in opposites
from 4 to 6 , they are moving in same directions
3. r '' (t) = 6t - 12
r ''(3) = 18-12 which is positive, so r ' (t) or the velocity of R is increasing.
4. when is 3t^2 - 12t + 9 = -2
3t^2 - 12t + 11 = 0
t = (12 ± √12)/6 = appr 2.577 or 1.423
both t = 2.577 and t = 1.423 fall in the interval 1 ≤ t ≤ 3