1. **Problem statement:** A rock of mass 3 kg slides down a rough plane inclined at 45° to the horizontal. It moves from point P to Q along the slope PQ = 37 m. The speed at P is 4 m/s, and the time taken to move from P to Q is 5.5 s. We need to find the speed at Q. 2. **Relevant formula:** We use the equation of motion for constant acceleration: $$v = u + at$$ where $v$ is the final velocity, $u$ is the initial velocity, $a$ is acceleration, and $t$ is time. 3. **Find acceleration:** We can also use the displacement formula: $$s = ut + \frac{1}{2}at^2$$ Given $s = 37$ m, $u = 4$ m/s, and $t = 5.5$ s, substitute: $$37 = 4 \times 5.5 + \frac{1}{2}a \times (5.5)^2$$ 4. Simplify: $$37 = 22 + \frac{1}{2}a \times 30.25$$ $$37 - 22 = 15.125a$$ $$15 = 15.125a$$ 5. Solve for $a$: $$a = \frac{15}{15.125}$$ $$a \approx 0.992$$ m/s² 6. **Calculate final velocity $v$:** $$v = 4 + 0.992 \times 5.5$$ $$v = 4 + 5.456$$ $$v = 9.456$$ m/s 7. **Round to 3 significant figures:** $$v \approx 9.46$$ m/s **Answer:** The speed of the rock at Q is approximately **9.46 m/s**. This matches closest to the option $v \approx 9.45$ m/s.