Suppose that the following force ~F(t) = 8>>><>>>: 2iˆ, t 2 [0, 1] −2iˆ, t 2 (1, 4] 0iˆ, t 2 (4, +¥) (1) acts on a particle of m = 1. Determine ~x(t),~v(t),~a(t) if x(t = 0) = 0, v(t = 0) = 0. Describe the story of this particle. In which times it accelerates/decelerates. Does it stop? If yes at what time?

To determine the position, velocity, and acceleration of the particle, we need to integrate the force function with respect to time.

1. For t in the interval [0, 1]:
Since F(t) = 2iˆ, the force remains constant within this interval. Thus, the equation of motion becomes:

m ~a(t) = ~F(t)
1 ~a(t) = 2iˆ
∫~a(t) dt = ∫2iˆ dt
~v(t) = 2tiˆ + ~C1 (where ~C1 is the constant of integration for velocity)

Now, applying the initial condition v(t = 0) = 0, we can solve for ~C1:
~v(t = 0) = 2(0)iˆ + ~C1 = ~C1 = 0
So, ~v(t) = 2tiˆ

Integrating again to find the position ~x(t):
∫~v(t) dt = ∫2tiˆ dt
~x(t) = t^2 iˆ + ~C2 (where ~C2 is the constant of integration for position)

Applying the initial condition x(t = 0) = 0, we can solve for ~C2:
~x(t = 0) = (0)^2 iˆ + ~C2 = ~C2 = 0
So, ~x(t) = t^2 iˆ

2. For t in the interval (1, 4]:
Since F(t) = -2iˆ, the force remains constant within this interval. The equation of motion becomes:

m ~a(t) = ~F(t)
1 ~a(t) = -2iˆ
∫~a(t) dt = ∫-2iˆ dt
~v(t) = -2t iˆ + ~C3 (where ~C3 is the constant of integration for velocity)

Applying the initial condition v(t = 0) = 0, we can solve for ~C3:
~v(t = 0) = -2(0)iˆ + ~C3 = ~C3 = 0
So, ~v(t) = -2t iˆ

Integrating again to find the position ~x(t):
∫~v(t) dt = ∫-2t iˆ dt
~x(t) = -t^2 iˆ + ~C4 (where ~C4 is the constant of integration for position)

Applying the initial condition x(t = 0) = 0, we can solve for ~C4:
~x(t = 0) = -(0)^2 iˆ+ ~C4 = ~C4 = 0
So, ~x(t) = -t^2 iˆ

3. For t in the interval (4, +∞):
Since F(t) = 0, there is no force acting on the particle, and hence, no acceleration.
Therefore, the velocity ~v(t) remains constant at -2tiˆ, and the position ~x(t) remains constant at -t^2iˆ.

Answer:
Position: ~x(t) = \begin{cases} t^2 i\hat{}, & \text{if } t \in [0, 1] \ -t^2 i\hat{}, & \text{if } t \in (1, 4] \ -t^2 i\hat{}, & \text{if } t \in (4, +\infty) \end{cases}
Velocity: ~v(t) = \begin{cases} 2t i\hat{}, & \text{if } t \in [0, 1] \ -2t i\hat{}, & \text{if } t \in (1, 4] \ -2t i\hat{}, & \text{if } t \in (4, +\infty) \end{cases}
Acceleration: ~a(t) = \begin{cases} 2 i\hat{}, & \text{if } t \in [0, 1] \ -2 i\hat{}, & \text{if } t \in (1, 4] \ 0 i\hat{}, & \text{if } t \in (4, +\infty) \end{cases}

The particle accelerates from t = 0 to t = 1 and decelerates from t = 1 to t = 4. After t > 4, it comes to a stop as there is no force acting on it.