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Part D dy Use the results from Parts B and C in the product rule of differentiation to find a simplified expression for the vertical velocity of the car, vy(t) = dt ► View Available Hint(s) vy(t) = Submit dy dt Part E yoe-at (cos(wt) + aw cos(wt)) -ayo²e-2at-w cos(wt) sin(wt) -yoeat (a cos(wt) + w sin(wt)) awy²e-2at cos(wt) sin(wt) Correct In this example we used both the chain and product rules of differentiation to obtain an expression for the (vertical) velocity-time relationship for the car from its position-time relationship. Previous Answers vy(0.25 s) = Evaluate the numerical value of the vertical velocity of the car at time t = 0.25 s using the expression from Part D, where yo = 0.75 m, a = 0.95 1/s, and w = 6.3 rad/s. ► View Available Hint(s) μÀ Value Units ?

Part D dy Use the results from Parts B and C in the product rule of differentiation to find a simplified expression for the vertical velocity of the car, vy(t) = dt ► View Available Hint(s) vy(t) = Submit dy dt Part E yoe-at (cos(wt) + aw cos(wt)) -ayo²e-2at-w cos(wt) sin(wt) -yoeat (a cos(wt) + w sin(wt)) awy²e-2at cos(wt) sin(wt) Correct In this example we used both the chain and product rules of differentiation to obtain an expression for the (vertical) velocity-time relationship for the car from its position-time relationship. Previous Answers vy(0.25 s) = Evaluate the numerical value of the vertical velocity of the car at time t = 0.25 s using the expression from Part D, where yo = 0.75 m, a = 0.95 1/s, and w = 6.3 rad/s. ► View Available Hint(s) μÀ Value Units ?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Part D dy Use the results from Parts B and C in the product rule of differentiation to find a simplified expression for the vertical velocity of the car, vy(t): dt ► View Available Hint(s) vy(t) = Submit Part E dy dt -at (cos(wt) + aw cos(wt)) w cos(wt) sin(wt) -yoe-at (a cos(wt) + w sin(wt)) awy²e-2 cos(wt) sin(wt) Yoe -ayo²e-2at Previous Answers -2at Correct In this example we used both the chain and product rules of differentiation to obtain an expression for the (vertical) velocity-time relationship for the car from its position-time relationship. Evaluate the numerical value of the vertical velocity of the car at time t - = 0.25 s using the expression from Part D, where yo = 0.75 m, a = 0.95 1/s, and w = 6.3 rad/s. ► View Available Hint(s) μÀ Vy(0.25 s) = Value Units ?
Transcribed Image Text:Part D dy Use the results from Parts B and C in the product rule of differentiation to find a simplified expression for the vertical velocity of the car, vy(t): dt ► View Available Hint(s) vy(t) = Submit Part E dy dt -at (cos(wt) + aw cos(wt)) w cos(wt) sin(wt) -yoe-at (a cos(wt) + w sin(wt)) awy²e-2 cos(wt) sin(wt) Yoe -ayo²e-2at Previous Answers -2at Correct In this example we used both the chain and product rules of differentiation to obtain an expression for the (vertical) velocity-time relationship for the car from its position-time relationship. Evaluate the numerical value of the vertical velocity of the car at time t - = 0.25 s using the expression from Part D, where yo = 0.75 m, a = 0.95 1/s, and w = 6.3 rad/s. ► View Available Hint(s) μÀ Vy(0.25 s) = Value Units ?
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An engineer is considering possible trajectories to use for emergency descent of a lunar module from low moon orbit to the lunar surface and decides to investigate one for which the vertical component of velocity as a function of time is described by vy(t) = voe 0.49 km/s, to = 390 s, and σ = 65 s. U0 = Part F For this trajectory, what would the vertical component of acceleration for the module be at time tm = too = 325 s? Recall that acceleration is the derivative of velocity with respect to time. ► View Available Hint(s) 0 O μà dây (325 s) = Value dt Units ? (t-to)² 202 where
Transcribed Image Text:An engineer is considering possible trajectories to use for emergency descent of a lunar module from low moon orbit to the lunar surface and decides to investigate one for which the vertical component of velocity as a function of time is described by vy(t) = voe 0.49 km/s, to = 390 s, and σ = 65 s. U0 = Part F For this trajectory, what would the vertical component of acceleration for the module be at time tm = too = 325 s? Recall that acceleration is the derivative of velocity with respect to time. ► View Available Hint(s) 0 O μà dây (325 s) = Value dt Units ? (t-to)² 202 where
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