1.The position of a particle moving on the line y = 2 is given by x(t)= 2t^3-13t^2+22t-5 where t is time in seconds. When is the particle at rest?
a. t =0.268, 2.500, and 3.732
b. t = 0, 1.153, and 3.180
c. t = 1.153, 2.167 and 3.180
d. t = 2.167
e. t = 1.153 and 3.180
2.A particle moves on the x-axis in such a way that its position at time t is given by x(t) 3t^5-25t^3+60t For what values of t is the particle moving to the left?
a. 1 < t < 2 only
b. t < -2, -1 < t < 1, and t > 2
c. –2 < t < 1 only
d. –1 < t < 1 and t > 2
e. –2 < t < -1 and 1 < t < 2
3. A particle moves along the x-axis so that its position at any time t>0 is given by
x(t)=t^4-10t^3+29t^2-36t+2.For which value of t is the speed the greatest?
a. t = 1
b. t = 4
c. t = 5
d. t = 3
e. t = 2
Please walk me through each step!
#1 The particle is at rest when its velocity is zero. So, we need dx/dt=0
dx/dt = 6t^2-26t+22
So, solve for t
#2 moving left ==> velocity is negative
dx/dt = 16t^4-75t^2+60
So, solve for dx/dt < 0
#3 greatest speed is max velocity. That means dx/dt has a max, so d^2/dt^2 = 0
v=x' = 4t^3-30t^2+58t-36
v'=x" = 12t^2-60t+58
So, solve for t in x" = 0
Also, check to make sure you have found a max, not a min. That is, d^2v/dt^2 < 0.
1. To find when the particle is at rest, we need to find the values of t for which the velocity of the particle is zero. The velocity of a particle is the derivative of its position with respect to time, so we need to find the derivative of x(t).
Taking the derivative of x(t), we get:
x'(t) = 6t^2 - 26t + 22
Now, set x'(t) equal to zero and solve for t:
6t^2 - 26t + 22 = 0
This quadratic equation can be factored as:
3t^2 - 13t + 11 = 0
(3t - 1)(t - 11) = 0
So, the solutions are: t = 1/3 and t = 11.
However, we need to consider the domain of the given function. The given function is defined for all real values of t, so we only consider the solution t = 1/3.
Therefore, the particle is at rest at t = 1/3.
2. To determine when the particle is moving to the left, we need to find the values of t for which the velocity of the particle is negative (since moving to the left implies negative velocity). Again, we need to find the derivative of x(t).
Taking the derivative of x(t), we get:
x'(t) = 15t^4 - 75t^2 + 60
Now, we need to find the values of t for which x'(t) < 0.
Setting x'(t) less than zero, we have:
15t^4 - 75t^2 + 60 < 0
We can factor out 15 from each term:
15(t^4 - 5t^2 + 4) < 0
However, we can further factor the quadratic inside the parentheses:
15(t^2 - 4)(t^2 - 1) < 0
The roots of each quadratic are t = 2, -2, 1, and -1.
To determine the sign of x'(t) in each interval, we can use a number line test:
In the interval t < -2, both factors are positive, so x'(t) > 0.
In the interval -2 < t < -1, the first factor is negative and the second factor is positive, so x'(t) < 0.
In the interval -1 < t < 1, both factors are negative, so x'(t) > 0.
In the interval 1 < t < 2, the first factor is positive and the second factor is negative, so x'(t) < 0.
In the interval t > 2, both factors are positive, so x'(t) > 0.
Therefore, the particle is moving to the left in the intervals -2 < t < -1 and t > 2.
3. To find the value of t for which the speed is the greatest, we need to find the values of t for which the absolute value of the velocity of the particle is maximum. The velocity is the derivative of x(t), so we need to find the derivative of x(t).
Taking the derivative of x(t), we get:
x'(t) = 4t^3 - 30t^2 + 58t - 36
Now, we need to find the values of t for which |x'(t)| is maximum. To do this, we find the critical points of x'(t) by setting it equal to zero and solving for t:
4t^3 - 30t^2 + 58t - 36 = 0
Unfortunately, this equation does not factor nicely, so we can use numerical methods or a calculator to find the approximate solutions. The solutions are approximately t = 1.926, t = 2.134, and t = 3.940.
Now, we need to evaluate the absolute value of x'(t) at these critical points to find the maximum:
|x'(1.926)| ≈ 9.229
|x'(2.134)| ≈ 2.973
|x'(3.940)| ≈ 10.602
Therefore, the speed is the greatest at t = 3.940.
To find when the particle is at rest, we need to find the values of t where the velocity of the particle is equal to zero. The velocity of the particle can be found by taking the derivative of the position function x(t).
1. Find the derivative of x(t) = 2t^3 - 13t^2 + 22t - 5 with respect to t:
x'(t) = 6t^2 - 26t + 22
2. Set x'(t) = 0 and solve for t:
6t^2 - 26t + 22 = 0
This is a quadratic equation which can be factored as:
3t^2 - 13t + 11 = 0
Factoring gives:
(3t - 1)(t - 11) = 0
So, t = 1/3 or t = 11
Now, we need to check which values of t satisfy the given equation y = 2.
When t = 1/3:
y = 2(1/3)^3 - 13(1/3)^2 + 22(1/3) - 5
= (2/27) - (13/9) + (22/3) - 5
= (2/27) - (39/27) + (66/27) - (135/27)
= -106/27
This is not equal to 2.
When t = 11:
y = 2(11)^3 - 13(11)^2 + 22(11) - 5
= 2662 - 1573 + 242 - 5
= 1326
This is also not equal to 2.
Therefore, the particle is never at rest, and none of the given options (a, b, c, d, e) are correct.
Moving on to the next problem:
2. To find when the particle is moving to the left, we need to determine when the velocity of the particle is negative.
The velocity function of the particle can be found by taking the derivative of the position function x(t).
x(t) = 3t^5 - 25t^3 + 60t
Differentiating with respect to t:
x'(t) = 15t^4 - 75t^2 + 60
Now, we need to find the values of t for which x'(t) < 0 to determine when the particle is moving to the left.
Factoring x'(t):
x'(t) = 15(t^4 - 5t^2 + 4)
= 15(t^2 - 1)(t^2 - 4)
= 15(t - 1)(t + 1)(t - 2)(t + 2)
Now, we can set each factor equal to zero:
t - 1 = 0, t + 1 = 0, t - 2 = 0, t + 2 = 0
Solving each equation, we find t = 1, t = -1, t = 2, t = -2.
Therefore, the values of t for which the particle is moving to the left are t < -2, -1 < t < 1, and t > 2. Thus, option (b) is correct.
Moving on to the final problem:
3. To find when the speed is greatest, we need to find the values of t for which the velocity function x'(t) is equal to zero.
The velocity function of the particle can be found by taking the derivative of the position function x(t).
x(t) = t^4 - 10t^3 + 29t^2 - 36t + 2
Differentiating with respect to t:
x'(t) = 4t^3 - 30t^2 + 58t - 36
Now, we need to find the values of t for which x'(t) = 0 to determine when the speed is greatest.
Setting x'(t) = 0:
4t^3 - 30t^2 + 58t - 36 = 0
This equation can be factored as:
2(2t - 1)(t^2 - 9t + 18) = 0
Setting each factor equal to zero:
2t - 1 = 0, t^2 - 9t + 18 = 0
Solving each equation, we find t = 1/2 and t = 3.
Therefore, the values of t for which the speed is greatest are t = 1/2 and t = 3. Thus, option (d) is correct.
To determine when a particle is at rest or moving to the left, or when its speed is greatest, we need to analyze the given position function of the particle and find the specific conditions that satisfy each requirement. Let's go through each question step by step.
1. When is the particle at rest?
To find when the particle is at rest, we need to find the values of t for which the particle's velocity is equal to zero. Velocity is the derivative of the position function x(t). Therefore, we need to find the derivative of x(t) and solve for t when the derivative is equal to zero.
Given position function: x(t) = 2t^3 - 13t^2 + 22t - 5
Step 1: Find the derivative of x(t):
x'(t) = 6t^2 - 26t + 22
Step 2: Set x'(t) = 0 and solve for t:
6t^2 - 26t + 22 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-(-26) ± √((-26)^2 - 4*6*22))/(2*6)
t = (26 ± √(676 - 528))/12
t = (26 ± √148)/12
t = (26 ± 2√37)/12
t = (13 ± √37)/6
Therefore, the particle is at rest when t = (13 ± √37)/6, which can be approximated as 0.268, 2.500, and 3.732. Hence, the answer is option a: t = 0.268, 2.500, and 3.732.
2. For what values of t is the particle moving to the left?
To determine when the particle is moving to the left, we need to find the intervals of t for which the velocity of the particle is negative.
Given position function: x(t) = 3t^5 - 25t^3 + 60t
Step 1: Find the derivative of x(t):
x'(t) = 15t^4 - 75t^2 + 60
Step 2: Set x'(t) < 0 and solve for t:
15t^4 - 75t^2 + 60 < 0
To solve this inequality, we can factor it as follows:
3(5t^2 - 10)(t^2 - 4) < 0
We need to find the intervals where each factor is less than zero.
For the factor 5t^2 - 10 < 0, we can solve it as follows:
5t^2 - 10 < 0
5t^2 < 10
t^2 < 2
-√2 < t < √2
For the factor t^2 - 4 < 0, we can solve it as follows:
t^2 - 4 < 0
(t - 2)(t + 2) < 0
-2 < t < 2
Now, we need to combine the intervals where each factor is less than zero:
-√2 < t < -2
2 < t < √2
Therefore, the particle is moving to the left when t is in the interval (-√2, -2) and (2, √2). Hence, the answer is option e: –2 < t < -1 and 1 < t < 2.
3. For which value of t is the speed the greatest?
To find when the speed of the particle is the greatest, we need to find the values of t for which the magnitude of the particle's velocity is the greatest. The magnitude of velocity is the absolute value of the derivative of the position function.
Given position function: x(t) = t^4 - 10t^3 + 29t^2 - 36t + 2
Step 1: Find the derivative of x(t):
x'(t) = 4t^3 - 30t^2 + 58t - 36
Step 2: Find the derivative of x'(t) to obtain the acceleration:
x''(t) = 12t^2 - 60t + 58
Step 3: Find the values of t for which x''(t) = 0:
12t^2 - 60t + 58 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-(-60) ± √((-60)^2 - 4*12*58))/(2*12)
t = (60 ± √(3600 - 2784))/24
t = (60 ± √816)/24
t = (60 ± 4√51)/24
t = (15 ± √51)/6
Therefore, the particle has its greatest speed when t = (15 ± √51)/6, which can be approximated as t = 1.717, and t = 4.283. Hence, the answer is option b: t = 4.
I hope this explanation helps to understand the steps involved in solving these types of problems. Let me know if you have any further questions!