Racing drivers experience large g-forces when driving fast round corners. One racing driver is driving a car in a circular ring which is 116 m in diameter. At one instant, his total acceleration is 12.6 m/s2 with a direction of 29 relative to the tangent.I think you have to draw a triangle and use Pythagoras to get ar and ata) At this instant,... i) what is the driver's radial acceleration? m/s2 ii) what is the driver's tangential acceleration? m/s2 b) Starting from the instant described above, and maintaining the same tangential acceleration, how long does it take the driver and car to complete one more complete lap? s(Hint: first find either initial angular or linear velocity.) EXERSCI 203 Formula SheetLinear KinematicsAngular KinematicsLinear KineticsAngular Kineticsv =স্বাধ স্বাধAx x-xiΔθ θα θεW =ΣΕ = maΔεΣΜ = Ιαtf-tiAttf-tiΔνV₁ - ViΔωW₁ - Winoxno a fā =Atv = v₁ + atAx = (0 +70%) t21no 4 Ax = vot + ₁₁₂at²w = w₁ + atA0 =α =F. At = m(v-vi)tf-tiΔεtf-tiM. At = I(wf-w₁)p = mv=(0-5000) tW = F.dH = IwW = M.01A0 =wot+at²E₁ = mg (Ah)M = Fd₁mv²Iw²no tv² = v² + 2a (Ax)w² = w² + 2α (A0)EkEk =22ntV₁ =Xi+1-xi-12hkx²kᎾ 2s = reEe=Ee =22Xi-1-2x + xi+1a₁ =V₁ = rwP = FvP = Mwh²highest point(v₁ sin 0)²in proj motYapex Yi-a₁ = ra1 =2g- ΣI = ICM + md²i=1time of flighty-y₁ = (v₁ sin 0)t; +gt1vay =w²rquadraticformular|a Totall=Other KinematicsFFT

Question

Racing drivers experience large g-forces when driving fast round corners. One racing driver is driving a car in a circular ring which is 116 m in diameter. At one instant, his total acceleration is 12.6 m/s2 with a direction of 29  relative to the tangent.I think you have to draw a triangle and use Pythagoras to get ar and ata) At this instant,...  i) what is the driver's radial acceleration?    m/s2  ii) what is the driver's tangential acceleration?   m/s2 b) Starting from the instant described above, and maintaining the same tangential acceleration, how long does it take the driver and car to complete one more complete lap?   s(Hint: first find either initial angular or linear velocity.) EXERSCI 203 Formula SheetLinear KinematicsAngular KinematicsLinear KineticsAngular Kineticsv =স্বাধ স্বাধAx x-xiΔθ θα θεW =ΣΕ = maΔεΣΜ = Ιαtf-tiAttf-tiΔνV₁ - ViΔωW₁ - Winoxno a fā =Atv = v₁ + atAx = (0 +70%) t21no 4 Ax = vot + ₁₁₂at²w = w₁ + atA0 =α =F. At = m(v-vi)tf-tiΔεtf-tiM. At = I(wf-w₁)p = mv=(0-5000) tW = F.dH = IwW = M.01A0 =wot+at²E₁ = mg (Ah)M = Fd₁mv²Iw²no tv² = v² + 2a (Ax)w² = w² + 2α (A0)EkEk =22ntV₁ =Xi+1-xi-12hkx²kᎾ 2s = reEe=Ee =22Xi-1-2x + xi+1a₁ =V₁ = rwP = FvP = Mwh²highest point(v₁ sin 0)²in proj motYapex Yi-a₁ = ra1 =2g- ΣI = ICM + md²i=1time of flighty-y₁ = (v₁ sin 0)t; +gt1vay =w²rquadraticformular|a Totall=Other KinematicsFFT

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