Question-30875

Question Number 30875 by Tinkutara last updated on 27/Feb/18

Commented by ajfour last updated on 28/Feb/18

$${sorry},\:{its}\:{just}\:\left(\mathrm{3}\right)\:{that}\:{is}\:{incorrect}. \\ $$

Answered by ajfour last updated on 28/Feb/18

tan θ=(y/x)=3t ⇒ y^2 =9t^2 x^2 r^2 =x^2 +y^2 =4t^2 (1+9t^2 ) ⇒ x^2 +y^2 =((4y^2 )/(9x^2 ))(((x^2 +y^2 )/x^2 )) ⇒ y^2 =(9/4)x^4 or y=(3/2)x^2 ....(i) (considering the upward parabola) since y=3tx x = 2t , y = 6t^2 v_x = 2 , a_x =0 v_y = (dy/dt) = 12t ⇒ a_y = 12 initial speed = (√(v_(x0) ^2 +v_(y0) ^2 )) =2m/s initial rate of speeding is initial tangential acceleration Now at t=0 , v^� = 2i^� while a^� = 12j^� so no initial tangential acceleration. as a_t = ((a^� .v^� )/(∣v^� ∣)) = 0 at t=0 . (v^2 /ρ) = a_(normal) (ρ is radius of curvature) so ρ = (v^2 /a_(normal) ) at t=0 we have min. radius of curvature since v is minimum and a_(normal) the maximum ρ_(minimum) = (4/(12)) = (1/3) but ρ_(maximum) → ∞ since as time proceeds v keeps increasing while a_(normal ) keeps decreasing .

$$\mathrm{tan}\:\theta=\frac{{y}}{{x}}=\mathrm{3}{t}\:\:\:\:\Rightarrow\:\:{y}^{\mathrm{2}} =\mathrm{9}{t}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}{t}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{9}{t}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{9}{x}^{\mathrm{2}} }\left(\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\:\:\:{y}^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}}{x}^{\mathrm{4}} \:\:\:{or}\:\:{y}=\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \:\:\:\:….\left({i}\right) \\ $$$$\left({considering}\:{the}\:{upward}\:{parabola}\right) \\ $$$${since}\:\:{y}=\mathrm{3}{tx} \\ $$$$\:\:\:\:{x}\:=\:\mathrm{2}{t}\:\:\:,\:\:\:{y}\:=\:\mathrm{6}{t}^{\mathrm{2}} \\ $$$$\:\:\:{v}_{{x}} =\:\mathrm{2}\:\:\:,\:\:{a}_{{x}} =\mathrm{0} \\ $$$$\:\:\:{v}_{{y}} =\:\frac{{dy}}{{dt}}\:=\:\mathrm{12}{t}\:\:\:\:\Rightarrow\:\:\:{a}_{{y}} =\:\mathrm{12} \\ $$$$\:\:{initial}\:{speed}\:=\:\sqrt{{v}_{{x}\mathrm{0}} ^{\mathrm{2}} +{v}_{{y}\mathrm{0}} ^{\mathrm{2}} }\:=\mathrm{2}{m}/{s} \\ $$$${initial}\:{rate}\:{of}\:{speeding}\:{is} \\ $$$$\:\:\:{initial}\:{tangential}\:{acceleration} \\ $$$${Now}\:\:{at}\:\:{t}=\mathrm{0}\:,\:\bar {{v}}\:=\:\mathrm{2}\hat {{i}} \\ $$$${while}\:\:\bar {{a}}\:=\:\mathrm{12}\hat {{j}} \\ $$$${so}\:{no}\:{initial}\:{tangential}\:{acceleration}. \\ $$$${as}\:\:{a}_{{t}} =\:\frac{\bar {{a}}.\bar {{v}}}{\mid\bar {{v}}\mid}\:=\:\mathrm{0}\:\:{at}\:{t}=\mathrm{0}\:\:. \\ $$$$\frac{{v}^{\mathrm{2}} }{\rho}\:=\:{a}_{{normal}} \:\:\:\:\:\left(\rho\:{is}\:{radius}\:{of}\:{curvature}\right) \\ $$$${so}\:\:\:\rho\:=\:\frac{{v}^{\mathrm{2}} }{{a}_{{normal}} }\: \\ $$$${at}\:\:{t}=\mathrm{0}\:\:{we}\:{have}\:{min}.\:{radius}\:{of} \\ $$$${curvature}\:{since}\:\boldsymbol{{v}}\:{is}\:{minimum} \\ $$$${and}\:{a}_{{normal}} \:\:\:{the}\:{maximum} \\ $$$$\rho_{{minimum}} \:=\:\frac{\mathrm{4}}{\mathrm{12}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${but}\:\rho_{{maximum}} \:\rightarrow\:\infty\:\:\:{since}\:{as} \\ $$$${time}\:{proceeds}\:\:{v}\:{keeps}\:{increasing}\: \\ $$$${while}\:{a}_{{normal}\:} \:{keeps}\:{decreasing}\:. \\ $$

Commented by Tinkutara last updated on 01/Mar/18

Thank you very much Sir! I got the answer.

Facebook Tweet Pin

Leave a Reply Cancel reply