Question Number 29463 by NECx last updated on 09/Feb/18
$${An}\:{object}\:{undergoes}\:{constant} \\ $$$${acceleration}\:{after}\:{starting}\:{from} \\ $$$${rest}\:{and}\:{then}\:{travels}\:\mathrm{5}{m}\:{in}\:{the} \\ $$$${first}\:{second}.{Determine}\:{how}\:{far} \\ $$$${it}\:{will}\:{go}\:{in}\:{the}\:{next}\:{second}. \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left.\mathrm{5}{m}\:\:{b}\right)\mathrm{10}{m}\:{c}\right)\mathrm{20}{m}\:{d}\right)\mathrm{5}{m} \\ $$$$ \\ $$$$ \\ $$
Answered by mrW2 last updated on 09/Feb/18
$${s}=\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$$${s}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{a}×\mathrm{1}^{\mathrm{2}} =\mathrm{5}\:{m} \\ $$$$\Rightarrow{a}=\mathrm{10} \\ $$$${s}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{2}^{\mathrm{2}} =\mathrm{20}\:{m} \\ $$$$\Delta{s}_{\mathrm{2}−\mathrm{1}} =\mathrm{20}−\mathrm{5}=\mathrm{15}\:{m} \\ $$$$\left.\Rightarrow{answer}\:{a}\right)\:{is}\:{correct}. \\ $$