1. **Problem statement:** A car travels with constant acceleration, passing points A, B, and C on a straight road. Velocity at A is 14 m/s. Distances AB and BC are equal. Time for AB is 5 s, for BC is 3 s. Find expressions for AB and AC in terms of acceleration $a$, then show $a=4$ m/s$^2$. Also find speed at C. 2. **Formulas and rules:** For constant acceleration, distance $s$ in time $t$ with initial velocity $u$ and acceleration $a$ is given by: $$s = ut + \frac{1}{2}at^2$$ Velocity after time $t$ is: $$v = u + at$$ 3. **Distance AB:** Initial velocity at A is $u=14$ m/s, time $t=5$ s, acceleration $a$ unknown. $$AB = 14 \times 5 + \frac{1}{2} a \times 5^2 = 70 + \frac{25}{2}a = 70 + 12.5a$$ 4. **Distance BC:** Since AB = BC, distance BC is also $70 + 12.5a$. 5. **Velocity at B:** Using $v = u + at$, $$v_B = 14 + 5a$$ 6. **Distance BC using velocity at B:** Initial velocity for BC is $v_B$, time $3$ s, acceleration $a$. $$BC = v_B \times 3 + \frac{1}{2} a \times 3^2 = 3(14 + 5a) + \frac{9}{2}a = 42 + 15a + 4.5a = 42 + 19.5a$$ 7. **Equate distances AB and BC:** $$70 + 12.5a = 42 + 19.5a$$ Simplify: $$70 - 42 = 19.5a - 12.5a$$ $$28 = 7a$$ $$a = 4$$ 8. **Distance AC:** Since AC = AB + BC = $2 \times AB$, $$AC = 2(70 + 12.5 \times 4) = 2(70 + 50) = 2 \times 120 = 240$$ 9. **Speed at C:** Velocity at C after total time $8$ s (5 + 3) is $$v_C = 14 + 4 \times 8 = 14 + 32 = 46$$ **Final answers:** - Acceleration $a = 4$ m/s$^2$ - Speed at C is 46 m/s