Question

Part D 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) = View Available Hint(s) vy(t) = Owyo'e-21 -yde (a cos(wt) +w sin(at)) cos(wt) sin(wt) yoe- (cos(wt)+aw cos(wt)) O ayo’e-20t -w cos(wt) sin(wt) Submit Previous Answers ✓ 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 Part E 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 s-, and w = 6.3s-1 View Available Hint(s) μΑ ? (0.25 s) =

Answer 1

To find the vertical velocity of the car using the product rule of differentiation, substitute the values from Part B and C into the given expression and simplify. Then evaluate the simplified expression at t = 0.25 s to find the numerical value of the vertical velocity.

Step-by-step explanation:

To find the vertical velocity of the car, we need to use the product rule of differentiation. The given expression for the vertical velocity is vy(t) = y'o*e(-2t1) + (a*cos(wt) +w*sin(wt))

Now, plug in the values from Part B and C into the expression. Substituting yo = 0.75 m, a = 0.95 s⁻¹, and w = 6.3 s⁻¹, we can simplify the expression.

Finally, evaluate the simplified expression at t = 0.25 s to find the numerical value of the vertical velocity of the car.