Answer:
R (t) = 60 - 60 cos (6t)
Step-by-step explanation:
Given that:
R(t) = acos (bt) + d
at t= 0
R(0) = 0
0 = acos (0) + d
a + d = 0 ----- (1)
After [tex]\dfrac{\pi}{12}[/tex] seconds it reaches a height of 60 cm from the ground.
i.e
[tex]R ( \dfrac{\pi}{12}) = 60[/tex]
[tex]60 = acos (\dfrac{b \pi}{12}) +d --- (2)[/tex]
Recall from the question that:
At t = 0, R(0) = 0 which is the minimum
as such it is only when a is negative can acos (bt ) + d can get to minimum at t= 0
Similarly; 60 × 2 = maximum
R'(t) = -ab sin (bt) =0
bt = k π
here;
k is the integer
making t the subject of the formula, we have:
[tex]t = \dfrac{k \pi}{b}[/tex]
replacing the derived equation of k into R(t) = acos (bt) + d
[tex]R (\dfrac{k \pi}{b}) = d+a cos (k \pi)[/tex] [tex]= \left \{ {{a+d \ for \ k \ odd} \atop {-a+d \ for k \ even}} \right.[/tex]
Since we known a < 0 (negative)
then d-a will be maximum
d-a = 60 × 2
d-a = 120 ----- (3)
Relating to equation (1) and (3)
a = -60 and d = 60
∴ R(t) = 60 - 60 cos (bt)
Similarly;
For [tex]R ( \dfrac{\pi}{12})[/tex]
[tex]R ( \dfrac{\pi}{12}) = 60 -60 \ cos (\dfrac{\pi b}{12}) =60[/tex]
where ;
[tex]cos (\dfrac{\pi b}{12}) =0[/tex]
Then b = 6
∴
R (t) = 60 - 60 cos (6t)
