To solve this problem, we need to find the final velocity of the particle after $t = 6$ seconds, given the force-time graph and the initial conditions (mass $m = 10 \, \text{kg}$, initial velocity $v_0 = 0$).

Steps to solve:

1. Use impulse-momentum principle: The change in momentum is equal to the impulse provided by the force.

   $\Delta p = m \Delta v = \text{Impulse} = \int F(t) \, dt$

   The area under the force-time graph gives the impulse. The total impulse will give the change in velocity since the particle starts from rest.

2. Break the graph into two parts:

   • From $t = 0$ to $t = 3$ seconds, the force is constant at $100 \, \text{N}$.
   • From $t = 3$ to $t = 6$ seconds, the force is constant at $-40 \, \text{N}$.

Calculations:

1. For $t = 0$ to $t = 3$:

   • Force $F = 100 \, \text{N}$.
   • Impulse $I_1 = F \times \Delta t = 100 \times 3 = 300 \, \text{Ns}$.

2. For $t = 3$ to $t = 6$:

   • Force $F = -40 \, \text{N}$.
   • Impulse $I_2 = (-40) \times (6 - 3) = -40 \times 3 = -120 \, \text{Ns}$.

3. Total impulse: $I_{\text{total}} = I_1 + I_2 = 300 + (-120) = 180 \, \text{Ns}$

4. Change in momentum: $\Delta p = m \Delta v = 180 \, \text{Ns}$ So, the change in velocity: $\Delta v = \frac{180}{m} = \frac{180}{10} = 18 \, \text{m/s}$

Since the initial velocity $v_0 = 0$, the final velocity $v_f$ is:

$v_f = 18 \, \text{m/s}$

Conclusion:

The final velocity of the particle at $t = 6$ seconds is $\boxed{18 \, \text{m/s}}$.

Would you like more details or have any questions?

Here are 5 related questions to further explore:

1. What is the physical significance of impulse in mechanics?
2. How does the shape of a force-time graph affect the motion of an object?
3. What happens if the force applied is variable instead of constant?
4. Can you derive the velocity using the kinematic equations without impulse?
5. How would the result change if the initial velocity was not zero?

Tip: Always check the area under the force-time graph when dealing with problems involving impulse!