Find the distance traveled in 45 seconds by an object traveling at a velocity of v(t) = 20 + 2cos(t) feet per second.
distance expression = 20t + 2sin(t) + c , where c is a constant
I will assume that you mean the distance in the first 45 seconds.
position at t=0
= 0 + 2sin0 + c = c
position at t = 45
= 900 + 2sin45 + c
distance travelled = 90+2sin45 + c - c
= appr. 901.7 feet
( notice in the second 45 seconds, the object would have travelled
1800 + 2sin90 + c - 900 - 2sin45 - x
= appr. 900.1 , different from the distance covered in the first 45 seconds.
If you wanted the actual distance of the path along the sine curve of the object you would have a much more difficult question)
To find the distance traveled in 45 seconds by an object with a given velocity function, we need to integrate the velocity function over the given time interval.
In this case, the velocity function is given by v(t) = 20 + 2cos(t) feet per second. To integrate this function, we follow these steps:
1. Determine the indefinite integral of the velocity function v(t) with respect to time t:
∫v(t) dt = ∫(20 + 2cos(t)) dt
The integral of a constant term is the product of the constant and the variable, and the integral of cos(t) is sin(t):
∫v(t) dt = 20t + 2∫cos(t) dt
= 20t + 2sin(t) + C
2. Evaluate the definite integral over the time interval [a, b]. In this case, we want to find the distance traveled in 45 seconds, so the time interval is [0, 45]:
∫[0,45]v(t) dt = [20t + 2sin(t)] from 0 to 45
= (20*45 + 2sin(45)) - (20*0 + 2sin(0))
= 900 + 2sin(45) - 0 - 2sin(0)
≈ 900 + 2(0.707) - 0 - 2(0)
≈ 901.414 feet
Therefore, the distance traveled in 45 seconds by an object with the given velocity function is approximately 901.414 feet.