A force-time graph is given. Calculate the impulse applied to an object and its change in momentum. The graph shows a force of 5 N applied from 0-3s (triangular), 3-5s (rectangular), and 5-8s (triangu

The graph can be divided into three geometric shapes:

• Area 1 (0 s to 3 s): A triangle. $$\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (3 \text{ s} - 0 \text{ s}) \times 5 \text{ N} = \frac{1}{2} \times 3 \text{ s} \times 5 \text{ N} = 7.5 \text{ N}\cdot\text{s}$$
• Area 2 (3 s to 5 s): A rectangle. $$\text{Area}_2 = \text{width} \times \text{height} = (5 \text{ s} - 3 \text{ s}) \times 5 \text{ N} = 2 \text{ s} \times 5 \text{ N} = 10 \text{ N}\cdot\text{s}$$
• Area 3 (5 s to 8 s): A triangle. $$\text{Area}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (8 \text{ s} - 5 \text{ s}) \times 5 \text{ N} = \frac{1}{2} \times 3 \text{ s} \times 5 \text{ N} = 7.5 \text{ N}\cdot\text{s}$$

The total impulse is the sum of these areas:

$$\text{Total Impulse} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = 7.5 \text{ N}\cdot\text{s} + 10 \text{ N}\cdot\text{s} + 7.5 \text{ N}\cdot\text{s} = 25 \text{ N}\cdot\text{s}$$

The mass of the particle is given as $m = 5 \text{ kg}$.

Given $v_i = 0 \text{ m/s}$.
Using the impulse-momentum theorem:

$$\text{Impulse} = m(v_f - v_i)$$ $$25 \text{ N}\cdot\text{s} = 5 \text{ kg} \times (v_f - 0 \text{ m/s})$$ $$25 = 5 v_f$$ $$v_f = \frac{25}{5} = 5 \text{ m/s}$$

Given $v_i = -4 \text{ m/s}$.
Using the impulse-momentum theorem:

$$\text{Impulse} = m(v_f - v_i)$$ $$25 \text{ N}\cdot\text{s} = 5 \text{ kg} \times (v_f - (-4 \text{ m/s}))$$ $$25 = 5 (v_f + 4)$$

Divide both sides by 5:

$$5 = v_f + 4$$ $$v_f = 5 - 4 = 1 \text{ m/s}$$

The final velocities are $5 \text{ m/s}$ when initially at rest, and $1 \text{ m/s}$ when initially at $-4 \text{ m/s}$.

Comparing these results with the given options, option A matches.

The final answer is $\text{A}$.