Question

Lamaj is rides his bike over a piece of gum and continues riding his bike at a constant rate time = 1.25 seconds the game is at a maximum height above the ground and one second later the Commons on the ground again

Answer 1

Lamaj rides his bike over a piece of gum and continues riding his bike at a constant rate. At time = 1.25 seconds, the gum is at a maximum height above the ground and 1 second later the gum is on the ground again.

a. If the diameter of the wheel is 68 cm, write an equation that models the height of the gum in centimeters above the ground at any time, t, in seconds.

b. What is the height of the gum when Lamaj gets to the end of the block at t = 15.6 seconds?

c. When are the first and second times the gum reaches a height of 12 cm?

Answer:

Step-by-step explanation:

a)

We are being told that:

Lamaj rides his bike over a piece of gum and continues riding his bike at a constant rate. This keeps the wheel of his bike in Simple Harmonic Motion and the Trigonometric equation  that models the height of the gum in centimeters above the ground at any time, t, in seconds.  can be written as:

$\mathbf {y = 34cos (\pi (t-1.25))+34}$

where;

y =  is the height of the gum at a given time (t) seconds

34 = amplitude of the motion

the amplitude of the motion was obtained by finding the middle between the highest and lowest point on the cosine graph.

$\mathbf{ \pi}$ = the period of the graph

1.25 = maximum vertical height stretched by 1.25 m  to the horizontal

b) From the equation derived above;

if we replace t with 1.56 seconds ; we can determine the height of the gum when Lamaj gets to the end of the block .

So;

$\mathbf {y = 34cos (\pi (15.6-1.25))+34}$

$\mathbf {y = 34cos (\pi (14.35))+34}$

$\mathbf {y = 34cos (45.08)+34}$

$\mathbf{y = 58.01}$

Thus, the  gum is at 58.01 cm from the ground at  t = 15.6 seconds.

c)

When are the first and second times the gum reaches a height of 12 cm

This indicates the position of y; so y = 12 cm

From the same equation from (a); we have :

$\mathbf {y = 34 cos(\pi (t-1.25))+34}$

$\mathbf{12 = 34 cos ( \pi(t-1.25))+34}$

$\dfrac {12-34}{34} = cos (\pi(t-1.25))$

$\dfrac {-22}{34} = cos(\pi(t-1.25))$

$2.27 = (\pi (t-1.25)$

t = 2.72 seconds

Similarly, replacing cosine in the above equation with sine; we have:

$\mathbf {y = 34 sin (\pi (t-1.25))+34}$

$\mathbf{12 = 34 sin ( \pi(t-1.25))+34}$

$\dfrac {12-34}{34} = sin (\pi(t-1.25))$

$\dfrac {-22}{34} = sin (\pi(t-1.25))$

$-0.703 = (\pi(t-1.25))$

t = 2.527 seconds

Hence, the gum will reach 12 cm first at 2.527 sec and second time at 2.72 sec.