At time t, the position of a body moving along the y-axis is y(t)=t^3-6t^2+9t.?

Question

a. find expressions for the velocity and acceleration of the moving body.

b. find the body's speed each time the acceleration is zero.

c. find the total distance traveled by the body from t=0 to t=2

Thank you very much.

Answer 1

a. The velocity dy/dt is the 1st derivative of t^3 - 6t^2 + 9t = 3t^2 - 12t + 9
The acceleration d^2y/dt^2 is the 2nd derivative of t^3 - 6t^2 + 9t = the 1st derivative of 3t^2 - 12t + 9 = 6t - 12
b. Set 6t - 12 = 0; t = 2; apply to the velocity equation.
c. This is trickier, since we're solving for distance not displacement. Solve y = t^3 - 6t^2 + 9t for the end points, t=0 and t=2. If the answer were displacement you'd just take the difference between these two values. But there may be one or two reversals of slope in that interval indicating a minimum or maximum, and this point may lie above or below both of the end-point values. So you need to find the y values at those points. At each reversal the 1st derivative dy/dt = 0. So you solve the dy/dt equation in quadratic form (3t^2 - 12t + 9 = 0) for roots (values of t) between 0 and 2. There will in fact be one. Solve for y at that point. The total distance will be the sum of two parts; the first from y at t=0 to y at the zero-velocity point, and the second from y at the zero velocity point to y at t=2. Add their absolute values.

Answer 2

a) V = ∂y/∂t, a = ∂V/∂t

b) Set a = 0 and use the resulting values of t in the eq for V

c) Y = y(2) - y(0)

You're welcome......