Question

KC has a piece of gum stuck to her bike tire, which applies a forward torque (i.e. a force to roll forwards) on the tire's movement. The torque varies in a periodic way that can be modeled approximately by a trigonometric function. When the gum is on the front of the tire, its weight is pulling forwards with a maximum torque of 0.01\text{ Nm}0.01 Nm0, point, 01, start text, space, N, m, end text (Newton metre, the SI unit for torque), and when it's on the back of the tire, it's pulling backwards with a minimum torque of -0.01\text{ Nm}−0.01 Nmminus, 0, point, 01, start text, space, N, m, end text. The maximum torque is reached once in every rotation, which is every 1.21.21, point, 2 meters. The first time it reaches its maximum torque is 0.30.30, point, 3 meters into the race.
Find the formula of the trigonometric function that models the torque \tauτtau the gum applies on the tire ddd meters into the race. Define the function using radians.

Answer 1

Answer:

  t(d) = 0.01cos(5π(d-0.3)/3)

Step-by-step explanation:

Since we are given the location of a maximum, it is convenient to use a cosine function to model the torque. The horizontal offset of the function will be 0.3 m, and the horizontal scaling will be such that one period is 1.2 m. The amplitude is given as 0.01 Nm.

The general form is ...

  torque = amplitude × cos(2π(d -horizontal offset)/(horizontal scale factor))

We note that 2π/1.2 = 5π/3. Filling in the given values, we have ...

  t(d) = 0.01·cos(5(d -0.3)/3)

Answer 2

Answer:

.01cos(2pi/1.2*(d-.3))

Step-by-step explanation: