The equation h=7 cos{pi/3 t} models the height h in centimeters after t seconds of a weight attached to the end of a spring that has been stretched an then released.
(C).Find the times at which the weight is at a height of 1 cm,of 3 cm,and of 5 cm below the rest position for the second time .Round your answers to the nearest hundredth.
for h=1,
1 = 7 cos(pi/3 t)
cos(pi/3 t) = 1/7
pi/3 t = 1.427
t = 1.363
and so on for the others
To find the times at which the weight is at a height of 1 cm, 3 cm, and 5 cm below the rest position, we need to set up the equation and solve for t. The equation we are given is:
h = 7cos(π/3t)
For the weight to be at a certain height, we substitute that height for h and solve for t.
For a height of 1 cm below the rest position:
1 = 7cos(π/3t)
To isolate t, we divide both sides by 7 and take the inverse cosine (or arc cosine) of both sides:
cos^(-1)(1/7) = π/3t
Now we can solve for t:
t = (3/cos^(-1)(1/7)) * π
Using a calculator, the value of cos^(-1)(1/7) is approximately 1.449. Plugging this value into the equation, we have:
t ≈ (3/1.449) * π
t ≈ 6.542 seconds (rounded to the nearest hundredth)
For a height of 3 cm below the rest position, we repeat the same process:
3 = 7cos(π/3t)
t = (3/cos^(-1)(3/7)) * π
Using a calculator, the value of cos^(-1)(3/7) is approximately 0.878. Plugging this value into the equation, we have:
t ≈ (3/0.878) * π
t ≈ 10.82 seconds (rounded to the nearest hundredth)
For a height of 5 cm below the rest position, we repeat the process once more:
5 = 7cos(π/3t)
t = (3/cos^(-1)(5/7)) * π
Using a calculator, the value of cos^(-1)(5/7) is approximately 0.603. Plugging this value into the equation, we have:
t ≈ (3/0.603) * π
t ≈ 15.68 seconds (rounded to the nearest hundredth)
Therefore, the times at which the weight is at a height of 1 cm, 3 cm, and 5 cm below the rest position for the second time are approximately 6.54 seconds, 10.82 seconds, and 15.68 seconds, respectively.