Question

Suppose that f(t) is continuous and twice-differentiable for t≥0. Further suppose f

′′
(t)≥5 for all t≥0 and f(0)=f
′
(0)=0. Using the Racetrack Principle, what linear function g(t) can we prove is less than or equal to f
′
(t) (for t≥0 )? g(t)= Then, also using the Racetrack Principle, what quadratic function h(t) can we prove is less than or equal to than f(t) (for t≥0) ? h(t)= For both parts of this problem, be sure you can clearly

Answer 1

Using the Racetrack Principle, given f''(t) ≥ 5 and initial conditions f(0) = f'(0) = 0, the linear function g(t) that is less than or equal to f'(t) is g(t) = 5t. The quadratic function h(t) that is less than or equal to f(t) is h(t) = (5/2)t^2.

Step-by-step explanation:

Given that f(t) is continuous and twice differentiable for t ≥0, and f''(t) ≥ 5, we can use the Racetrack Principle to determine bounds on both f'(t) and f(t) starting from t = 0 where f(0) = f'(0) = 0. To find the linear function g(t) such that g(t) ≤ f'(t) we would integrate the second derivative from 0 to t, knowing that the acceleration (f''(t)) ≥ 5. This gives us a minimum velocity function g(t) = 5t, since integrating the constant acceleration 5 from 0 to t results in 5t. The Racetrack Principle suggests that since f'(0) = 0 and f''(t) ≥ 5, f'(t) must be at least as large as its minimum value at t = 0 plus the integral of 5 from 0 to t, which yields g(t) = 5t.

To determine h(t), we integrate g(t) to get the minimum distance function such that h(t) ≤ f(t). The integration of g(t) = 5t from 0 to t gives us h(t) = (5/2)t^2. This means h(t) is the quadratic function we find by applying the Racetrack Principle to the initial conditions and acceleration, suggesting that h(t) = (5/2)t^2 ≤ f(t).