JAMB Physics 2021 Paper

Concept:The work done by the engine appears as the car's gain in kinetic energy, and power is the rate of doing that work.Explanation:The car starts from rest, so its initial kinetic energy is $0$.Final kinetic energy is $E_k = \frac{1}{2}mv^2$.Substitute $m = 800\, \text{kg}$ and $v = 25\, \text{m/s}$:$E_k = \frac{1}{2} \times 800 \times (25)^2$$E_k = 400 \times 625 = 250{,}000\, \text{J}$.This equals the work done by the engine in $t = 20\, \text{s}$.Power is $P = \frac{\text{work}}{\text{time}} = \frac{E_k}{t}$.So $P = \frac{250{,}000}{20} = 12{,}500\, \text{W}$.In standard form, $P = 1.25 \times 10^4\, \text{W}$.Do not forget the $\frac{1}{2}$ factor; using only $\frac{mv^2}{t}$ would give twice the correct power.Answer:A. $1.25 \times 10^4\, \text{W}$