A particle moves along the x-axis in such a way that it's position in time t for t is greator or equal to 0 is given by x= 1/3t^3 - 3t^2 +8
A) show that at time t= 0 the particle is moving to the right.
B) find all values of t for which the particle is moving to the left.
C) what is the position of the particle at time t=3
D) when t=3, what is the total distance the particle has traveled
To answer these questions, we need to analyze the given position function, x = (1/3)t^3 - 3t^2 + 8, which describes the particle's motion along the x-axis.
A) To determine whether the particle is moving to the right or left at time t = 0, we need to evaluate the derivative of x with respect to time (dx/dt) at t = 0. If dx/dt > 0, the particle is moving to the right. If dx/dt < 0, the particle is moving to the left.
To find the derivative, we can differentiate the position function with respect to time:
dx/dt = d/dt [(1/3)t^3 - 3t^2 + 8]
dx/dt = (1/3)(3t^2) - 2(3t)
dx/dt = t^2 - 6t
Now, evaluate dx/dt at t = 0:
dx/dt = (0)^2 - 6(0) = 0
Since dx/dt at t = 0 is equal to 0, the particle is not moving at that instant. Therefore, it is neither moving to the right nor to the left.
B) To find the values of t for which the particle is moving to the left, we need to determine when dx/dt is negative (dx/dt < 0). Solve the inequality t^2 - 6t < 0.
Factor the expression by factoring out t: t(t - 6) < 0
To determine when the inequality is true, we need to analyze the signs of t and t - 6. Considering three intervals: (-∞, 0), (0, 6), and (6, ∞), we can see that the inequality holds true for t in the interval (0, 6).
Therefore, the values of t for which the particle is moving to the left are t ∈ (0, 6).
C) To find the position of the particle at time t = 3, substitute t = 3 into the position function:
x = (1/3)(3^3) - 3(3^2) + 8
x = 9 - 27 + 8
x = -10
The position of the particle at time t = 3 is -10.
D) To find the total distance traveled by the particle when t = 3, we need to find the sum of the absolute values of the distances covered while moving to the right and to the left.
Since we established earlier that the particle is not moving at t = 0, it starts moving at t > 0.
The distance traveled while moving to the right is given by the integral of dx from t = 0 to t = 3:
∫[0,3] (t^2 - 6t) dt
Integrating this expression yields:
[-(1/3)t^3 + 3t^2] [0,3]
= [-(1/3)(3^3) + 3(3^2)] - [-(1/3)(0^3) + 3(0^2)]
= -(1/3)(27) + 3(9)
= -9 + 27
= 18
The distance traveled while moving to the left is also found by evaluating the integral of dx from t = 0 to t = 3, but with the absolute value of dx/dt:
∫[0,3] |t^2 - 6t| dt
Integrating this expression yields:
∫[0,3] (6t - t^2) dt
= [3t^2 - (1/3)t^3] [0,3]
= 3(3^2) - (1/3)(3^3) - [3(0^2) - (1/3)(0^3)]
= 27 - 9 - 0
= 18
Therefore, the total distance traveled by the particle when t = 3 is 18 + 18 = 36 units.