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
= 3(t-1)(t-3)
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
YEAHHH
To answer your questions, let's go step by step:
1. To find the times during which particle R is moving to the left, we need to determine when its velocity is negative.
The velocity of particle R is given by the derivative of its position function:
v(t) = dr(t)/dt = 3t^2 - 12t + 9
To find when v(t) is negative, we need to solve the inequality:
3t^2 - 12t + 9 < 0
To do that, we can find the roots of the quadratic equation:
3t^2 - 12t + 9 = 0
By factoring, we get:
3(t-1)(t-3) = 0
The roots of this equation are t = 1 and t = 3.
Therefore, particle R is moving to the left for 0 ≤ t < 1 and 3 < t ≤ 6.
2. To find the times during which the two particles travel in opposite directions, we need to determine when their velocities have opposite signs.
The velocity of particle P is given by the derivative of its position function:
v(t) = dp(t)/dt = -2πsin((π/4)t)
For v(t) and vR(t) to have opposite signs, one must be positive while the other is negative.
As v(t) = -2πsin((π/4)t), we can see that it is negative for 0 < t < 4 and positive for 4 < t ≤ 6.
Therefore, the two particles travel in opposite directions for 4 < t ≤ 6.
3. To find the acceleration of particle P at t = 3, we need to find the second derivative of its position function:
a(t) = d^2p(t)/dt^2 = -2π^2*cos((π/4)t)
By substituting t = 3 into the acceleration function, we get:
a(3) = -2π^2*cos((π/4) * 3) = -2π^2*cos(3π/4) = -2π^2*(-√2/2) = π^2√2
The acceleration of particle P at t = 3 is π^2√2.
Since the acceleration is positive (π^2 > 0) at t = 3, particle P is speeding up.
4. To show that particle R must have a velocity of -2 at some point during the interval (1, 3), we need to find the velocity function v(t) and evaluate it at t = 1 and t = 3.
The velocity of particle R is given by the derivative of its position function:
v(t) = dr(t)/dt = 3t^2 - 12t + 9
Evaluate v(t) at t = 1 and t = 3:
v(1) = 3(1)^2 - 12(1) + 9 = 3 - 12 + 9 = 0
v(3) = 3(3)^2 - 12(3) + 9 = 27 - 36 + 9 = 0
From the calculations, it appears that the velocity of particle R is 0 at t = 1 and t = 3.
Therefore, particle R must have a velocity of -2 at some point during the interval (1, 3).
This is because the velocity changes sign from positive to negative, meaning it must pass through -2 at some point within this interval.
To solve these questions, we'll need to analyze the position, velocity, and acceleration of particles P and R during the given time interval.
1. To determine when particle R is moving to the left, we need to find the values of t for which its velocity, r'(t), is negative. The velocity of particle R is given by r'(t) = 3t^2 - 12t + 9. Set r'(t) < 0 and solve for t:
3t^2 - 12t + 9 < 0
Factorizing the quadratic equation, we get:
(t - 1)(t - 3) < 0
This inequality tells us that the expression is negative between t = 1 and t = 3. Therefore, particle R is moving to the left during the interval t = 1 to t = 3.
2. To find the times when the two particles travel in opposite directions, we need to determine when their velocities have opposite signs. Particle P's velocity is given by p'(t) = -((pi/4)sin((pi/4)t)), while particle R's velocity is given by r'(t) = 3t^2 - 12t + 9. Set p'(t) * r'(t) < 0 and solve for t:
-((pi/4)sin((pi/4)t))(3t^2 - 12t + 9) < 0
Now, we need to determine the sign of each factor individually.
For p'(t) = -((pi/4)sin((pi/4)t)), we know it changes sign at t = 2 and t = 6, where sin((pi/4)t) changes sign.
For r'(t) = 3t^2 - 12t + 9, we can find the roots by setting it equal to zero:
3t^2 - 12t + 9 = 0
(t - 1)(t - 3) = 0
This means it changes sign at t = 1 and t = 3.
To determine when p'(t) * r'(t) < 0, we can use a sign table:
t < 1 1 < t < 2 2 < t < 3 3 < t < 6 t > 6
p'(t) | - | + | - | + | +
r'(t) | - | - | + | - | +
From the sign table, we can see that p'(t) * r'(t) < 0 for t in the intervals 1 < t < 2 and 3 < t < 6. Therefore, the two particles travel in opposite directions during these time intervals.
3. To find the acceleration of particle P at t = 3, we need to take the derivative of its velocity, p'(t), which is p''(t). The acceleration is given by p''(t) = -((pi^2)/16)cos((pi/4)t).
Substituting t = 3 into p''(t), we get:
p''(3) = -((pi^2)/16)cos((pi/4)*3)
= -((pi^2)/16)cos((3pi/4))
To determine if particle P is speeding up, slowing down, or neither at t = 3, we need to analyze the sign of p''(3). Since the cosine function is negative in the second quadrant, we know that p''(3) is negative. Therefore, particle P is slowing down at t = 3.
4. To show that particle R must have a velocity of -2 during the interval (1,3), we need to find the values of t for which r'(t) = -2.
Setting r'(t) = -2, we have:
3t^2 - 12t + 9 = -2
Simplifying the equation:
3t^2 - 12t + 11 = 0
To determine if there is a solution in the interval (1,3), we can check if the discriminant is greater than zero:
Δ = b^2 - 4ac = (-12)^2 - 4(3)(11) = 144 - 132 = 12 > 0
Since the discriminant is positive, there must be at least one instant when particle R has a velocity of -2 during the interval (1,3).
By following the steps above, you can solve these questions related to the motion of particles P and R along the x-axis.