A weight is attached to to a spring that is fixed to the floor. The equation h=7cos (pi/3 t) models the height, h, in centimeters after t seconds of the weight being stretched and released.


Solve for the solution of t and find the times at which the weight is first at a height of 1 cm, of 3 cm, and of 5 cm above the rest position. Round to the nearest hundreth

Verified Answer :

lets solve for t in the height equation:

h = 7 cos (pi/3 t)
cos (pi/3 t) = h/7
pi/3 t = cos^-1 (h/7)

t = (3/pi) cos^-1 (h/7)

then we substitute:

t = (3/pi) cos^-1 (1/7)
t = 78.1 s

t = (3/pi) cos^-1 (3/7)
t= 61.71

t = (3/pi) cos^-1 (5/7)
t = 42.41

those are the times in seconds respectively.

Answer:

\(t=\dfrac{3}{\pi}\left(\cos ^{-1}\dfrac{h}{7}\right)\)

Step-by-step explanation:

The given function is,

\(h=7\cos \left(\dfrac{\pi}{3}t\right)\)

\(\Rightarrow \cos \left(\dfrac{\pi}{3}t\right)=\dfrac{h}{7}\)

\(\Rightarrow \dfrac{\pi}{3}t=\cos ^{-1}\dfrac{h}{7}\)

\(\Rightarrow t=\dfrac{3}{\pi}\left(\cos ^{-1}\dfrac{h}{7}\right)\)

This is the solution of t.

Now we have to find t when h is 1 cm, 3 cm and 5 cm.

When h=1

\(t=\dfrac{3}{\pi}\left(\cos ^{-1}\dfrac{1}{7}\right)=78.10\ s\)

When h=3

\(t=\dfrac{3}{\pi}\left(\cos ^{-1}\dfrac{3}{7}\right)=61.71\ s\)

When h=5

\(t=\dfrac{3}{\pi}\left(\cos ^{-1}\dfrac{5}{7}\right)=42.41\ s\)