Question Number 8609 by tawakalitu last updated on 17/Oct/16
Answered by ridwan balatif last updated on 18/Oct/16
Answered by ridwan balatif last updated on 18/Oct/16
Commented by tawakalitu last updated on 18/Oct/16
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$
Answered by sandy_suhendra last updated on 18/Oct/16
Commented by sandy_suhendra last updated on 18/Oct/16
$$\mathrm{let}\:\mathrm{the}\:\mathrm{sport}\:\mathrm{car}\:\mathrm{overtakes}\:\mathrm{the}\:\mathrm{car}\:\mathrm{at}\:\mathrm{t}\:\mathrm{second}\:\mathrm{after}\:\mathrm{the}\:\mathrm{car}\:\mathrm{passes}\:\mathrm{the}\:\mathrm{sport}\:\mathrm{car} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car}\:=\:\mathrm{the}\:\mathrm{area}\:\mathrm{below}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car}\:=\mathrm{20t}\:\mathrm{meter} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sport}\:\mathrm{car}=\mathrm{the}\:\mathrm{area}\:\mathrm{below}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sport}\:\mathrm{car}=\frac{\mathrm{t}−\mathrm{10}+\mathrm{t}−\mathrm{20}}{\mathrm{2}}×\mathrm{30}=\left(\mathrm{30t}−\mathrm{450}\right)\:\mathrm{meter} \\ $$$$\mathrm{the}\:\mathrm{sport}\:\mathrm{car}\:\mathrm{will}\:\mathrm{overtake}\:\mathrm{the}\:\mathrm{car}\:\mathrm{if} \\ $$$$\mathrm{20t}=\mathrm{30t}−\mathrm{450} \\ $$$$\:\:\:\:\:\mathrm{t}=\mathrm{45}\:\mathrm{second} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{distance}\:=\:\mathrm{20}×\mathrm{45}=\mathrm{900}\:\mathrm{meter} \\ $$
Commented by tawakalitu last updated on 18/Oct/16
$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$