Question Number 16505 by Tinkutara last updated on 23/Jun/17
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{magnitude}\:\mathrm{of} \\ $$$$\mathrm{change}\:\mathrm{in}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{motorcycle} \\ $$$$\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{speed}\:{v}_{\mathrm{0}} \:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{circular}\:\mathrm{path}\:\mathrm{of}\:\mathrm{length}\:{l}\:=\:\frac{\pi}{\mathrm{3}}{R}\:\mathrm{and} \\ $$$$\mathrm{radius}\:\mathrm{R}?\:\mathrm{Treat}\:\mathrm{motorcycle}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{particle} \\ $$$$\left(\mathrm{1}\right)\:\mid\Delta\overset{\rightarrow} {{v}}\mid\:=\:\frac{{v}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\mid\Delta\overset{\rightarrow} {{v}}\mid\:=\:{v}_{\mathrm{0}} \\ $$
Answered by ajfour last updated on 23/Jun/17
$$\mid\Delta\overset{\rightarrow} {{v}}\mid^{\mathrm{2}} =\:\mid\overset{\rightarrow} {{v}}_{{f}} −\overset{\rightarrow} {{v}}_{{i}} \mid^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:={v}_{{i}} ^{\mathrm{2}} +{v}_{{f}} ^{\mathrm{2}} −\mathrm{2}{v}_{{i}} {v}_{{f}} \:\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{v}_{\mathrm{0}} ^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} −\mathrm{2}{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{cos}\:\left(\pi/\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{v}_{\mathrm{0}} ^{\mathrm{2}} \: \\ $$$${therefore}\:\:{option}\:\:\left(\mathrm{2}\right)\:. \\ $$
Commented by ajfour last updated on 23/Jun/17
Commented by Tinkutara last updated on 24/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$