JAMB Physics 2023 Paper

Concept:Use the equation of uniformly accelerated motion to find the initial velocity, then compare distances covered in different time intervals.Explanation:Given: $a = 4\,\mathrm{m\,s}^{-2}$, $s = 250\,\mathrm{m}$, $t = 10\,\mathrm{s}$.Using $s = ut + \frac{1}{2}at^2$, substitute the values:$$250 = u(10) + \frac{1}{2}(4)(10)^2$$$$250 = 10u + 200$$$$10u = 50$$$$u = 5\,\mathrm{m\,s}^{-1}$$This is the initial velocity at the start of the first $10\,\mathrm{s}$.Now find the total distance covered in $20\,\mathrm{s}$:$$s_{20} = 5(20) + \frac{1}{2}(4)(20)^2$$$$s_{20} = 100 + 800 = 900\,\mathrm{m}$$The distance covered in the next $10\,\mathrm{s}$ is the distance from $t = 10\,\mathrm{s}$ to $t = 20\,\mathrm{s}$:$$= s_{20} - s_{10} = 900 - 250 = 650\,\mathrm{m}$$Answer:650 m (Option A).