Question Number 226558 by hardmath last updated on 04/Dec/25
Answered by mr W last updated on 04/Dec/25
Commented by mr W last updated on 04/Dec/25
$${v}_{\mathrm{1}} =\mathrm{4}×\mathrm{5}=\mathrm{20}\:{m}/{s}\:\:\left(\rightarrow\right) \\ $$$${d}_{\mathrm{1}} =\frac{\mathrm{4}×\mathrm{5}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{50}\:{m} \\ $$$$\:{t}={t}_{\mathrm{2}} −{t}_{\mathrm{1}} \\ $$$$−\mathrm{50}=\mathrm{20}\left(\:{t}\right)−\frac{\mathrm{4}\left(\:{t}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\:{t}=\mathrm{5}+\mathrm{5}\sqrt{\mathrm{2}} \\ $$$${v}_{\mathrm{2}} =\mathrm{20}−\mathrm{4}\left(\mathrm{5}+\mathrm{5}\sqrt{\mathrm{2}}\right)=−\mathrm{20}\sqrt{\mathrm{2}}\:{m}/{s}\:\:\left(\leftarrow\right) \\ $$