Find the distance traveled in 45 seconds by an object traveling at a velocity of v(t) = 20 + 3cos(t) feet per second.
s(t) = ∫v(t) dt = 20t + 3sin(t)
so plug in t=45
To find the distance traveled in 45 seconds by an object traveling at a given velocity function, we can use the concept of definite integrals. The distance traveled is obtained by finding the integral of the velocity function over the given time interval.
Step 1: Evaluate the Definite Integral
First, we need to find the definite integral of the velocity function over the given time interval. In this case, the integral represents the displacement of the object within the 45-second interval:
∫[0 to 45] (20 + 3cos(t)) dt
Step 2: Calculate the Integral
To evaluate the integral, we can integrate the velocity function term by term. The integral of a constant term (20) with respect to t yields (20t). The integral of cos(t) with respect to t is given by sin(t). Applying the Fundamental Theorem of Calculus, we have:
∫[0 to 45] (20 + 3cos(t)) dt = [20t + 3sin(t)] evaluated from t = 0 to t = 45
Step 3: Evaluate the Integral Limits
To determine the displacement within the 45-second interval, we substitute the upper limit (t = 45) and subtract the value at the lower limit (t = 0):
[20(45) + 3sin(45)] - [20(0) + 3sin(0)]
Simplifying this expression gives us the distance traveled in 45 seconds by the object.