The equation h=7 cos{pi/3 t} models the height h in centimetres after t seconds of a weight attached to the end of a spring that has been stretched and then released.
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 answer to the nearest hundredth.
To find the times at which the weight is at a specific height, we can substitute the given height values into the equation h=7cos(pi/3 t) and solve for t.
Let's start by finding the time when the weight is at a height of 1 cm below the rest position:
1 = 7cos(pi/3 t)
To isolate cos(pi/3 t), we can divide both sides by 7:
1/7 = cos(pi/3 t)
Now, we need to find the inverse cosine (also known as arccos) of both sides to solve for t:
arccos(1/7) = pi/3 t
To find the value of arccos(1/7), we can use a scientific calculator or a math software. Plugging in the value gives:
arccos(1/7) ≈ 1.479 radians
Now, we divide this value by pi/3 to solve for t:
1.479 / (pi/3) ≈ 1.479 * 3 / pi ≈ 1.41
So, the weight is at a height of 1 cm below the rest position at approximately t = 1.41 seconds.
Now, let's repeat the process for the heights of 3 cm and 5 cm:
For 3 cm below the rest position:
3 = 7cos(pi/3 t)
cos(pi/3 t) = 3/7
arccos(3/7) = pi/3 t
arccos(3/7) ≈ 0.877 radians
0.877 / (pi/3) ≈ 0.877 * 3 / pi ≈ 0.83
For 5 cm below the rest position:
5 = 7cos(pi/3 t)
cos(pi/3 t) = 5/7
arccos(5/7) = pi/3 t
arccos(5/7) ≈ 0.584 radians
0.584 / (pi/3) ≈ 0.584 * 3 / pi ≈ 0.55
Thus, the weight is at a height of 3 cm below the rest position at approximately t = 0.83 seconds, and at a height of 5 cm below the rest position at approximately t = 0.55 seconds.
Remember to round your answers to the nearest hundredth as specified in the question.
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 can equate each height value with the equation h = 7cos(π/3t) and solve for t.
Let's start with the height of 1 cm:
1 = 7cos(π/3t)
To solve for t, we need to isolate t on one side of the equation. Divide both sides by 7:
1/7 = cos(π/3t)
Now, take the inverse cosine of both sides to find the value of π/3t:
cos^(-1)(1/7) = π/3t
Using a calculator, find:
π/3t ≈ 1.346
Now divide both sides by π/3:
t ≈ 1.346 / (π/3)
t ≈ 1.346 * (3/π)
t ≈ 1.346 * 0.955
t ≈ 1.286 seconds (rounded to the nearest hundredth)
Therefore, the weight is at a height of 1 cm below the rest position for the second time at approximately 1.29 seconds.
Repeat the same steps for the heights of 3 cm and 5 cm:
For the height of 3 cm:
3 = 7cos(π/3t)
cos^(-1)(3/7) = π/3t
t ≈ 2.296 seconds (rounded to the nearest hundredth)
Therefore, the weight is at a height of 3 cm below the rest position for the second time at approximately 2.30 seconds.
For the height of 5 cm:
5 = 7cos(π/3t)
cos^(-1)(5/7) = π/3t
t ≈ 0.676 seconds (rounded to the nearest hundredth)
Therefore, the weight is at a height of 5 cm below the rest position for the second time at approximately 0.68 seconds.
To summarize, the times at which the weight is at heights of 1 cm, 3 cm, and 5 cm below the rest position for the second time are approximately:
- 1 cm: 1.29 seconds
- 3 cm: 2.30 seconds
- 5 cm: 0.68 seconds