a. To solve the equation h=7cos(π/3t) for t, divide both sides by 7 to isolate the cosine function:
h/7 = cos(π/3t)
Now, take the inverse cosine (arccos) of both sides to solve for t:
π/3t = arccos(h/7)
Now, divide by π/3 to solve for t:
t = (3/π) * arccos(h/7)
b. To 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, substitute the respective heights into the equation and solve for t:
1 cm:
t = (3/π) * arccos(1/7) ≈ 0.80 seconds
3 cm:
t = (3/π) * arccos(3/7) ≈ 1.40 seconds
5 cm:
t = (3/π) * arccos(5/7) ≈ 2.12 seconds
Therefore, the weight reaches a height of 1 cm at approximately 0.80 seconds, a height of 3 cm at approximately 1.40 seconds, and a height of 5 cm at approximately 2.12 seconds above the rest position.
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.....
1 answer