Question
A weight is attached to a spring that is fixed to the floor. The equation h=7cos (π3t) models the height, h, in centimeters after t seconds of the weight being stretched and released.a. Solve the equation for t.b. 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. Calculate your answer using radian angle measure. Round your answers to the nearest hundredth show work in short term
Answers
GPT 3.5
a. To solve the equation for t, we need to isolate t.
Given: h = 7cos(π/3t)
1. Divide both sides by 7:
h/7 = cos(π/3t)
2. Take the inverse cosine of both sides to isolate t:
cos^(-1)(h/7) = π/3t
3. Multiply both sides by 3/π to solve for t:
t = 3cos^(-1)(h/7)/π
b. To find the times at which the weight is first at a height of 1 cm, 3 cm, and 5 cm above the rest position, we substitute these values into the equation and solve for t.
1. For h = 1 cm:
t = 3cos^(-1)(1/7)/π ≈ 0.53 seconds
2. For h = 3 cm:
t = 3cos^(-1)(3/7)/π ≈ 1.51 seconds
3. For h = 5 cm:
t = 3cos^(-1)(5/7)/π ≈ 2.26 seconds
Therefore, the times at which the weight is first at a height of 1 cm, 3 cm, and 5 cm above the rest position are approximately 0.53 seconds, 1.51 seconds, and 2.26 seconds, respectively.
Given: h = 7cos(π/3t)
1. Divide both sides by 7:
h/7 = cos(π/3t)
2. Take the inverse cosine of both sides to isolate t:
cos^(-1)(h/7) = π/3t
3. Multiply both sides by 3/π to solve for t:
t = 3cos^(-1)(h/7)/π
b. To find the times at which the weight is first at a height of 1 cm, 3 cm, and 5 cm above the rest position, we substitute these values into the equation and solve for t.
1. For h = 1 cm:
t = 3cos^(-1)(1/7)/π ≈ 0.53 seconds
2. For h = 3 cm:
t = 3cos^(-1)(3/7)/π ≈ 1.51 seconds
3. For h = 5 cm:
t = 3cos^(-1)(5/7)/π ≈ 2.26 seconds
Therefore, the times at which the weight is first at a height of 1 cm, 3 cm, and 5 cm above the rest position are approximately 0.53 seconds, 1.51 seconds, and 2.26 seconds, respectively.