A particle moves with a velocity of v(t)=3cos(2t) m/s along an axis. Find the distance traveled by the particle over the time interval 0 is less than or equal to "t", which is less than or equal to (pi/8)
distance is just the integral of velocity. So, just integrate, and evaluate at the ends of the interval.
To find the distance traveled by the particle over the time interval 0 ≤ t ≤ π/8, we need to integrate the absolute value of the velocity function.
The velocity function is v(t) = 3cos(2t).
To calculate the distance traveled, we integrate the absolute value of the velocity function over the given time interval:
Distance = ∫ |v(t)| dt
Let's break down the calculation step by step:
Step 1: Find the absolute value of the velocity function:
|v(t)| = |3cos(2t)|
Step 2: Integrate the absolute value of the velocity function over the given time interval:
Distance = ∫ |3cos(2t)| dt
Step 3: Apply the properties of integration:
Distance = ∫ (3cos(2t)) dt
Step 4: Integrate the function:
Distance = 3∫ cos(2t) dt
To integrate cos(2t), we use the u-substitution method, where u = 2t and du = 2dt.
Step 5: Substitute the values for u and du:
Distance = (3/2)∫ cos(u) du
Step 6: Integrate cos(u):
Distance = (3/2)sin(u) + C
Step 7: Substitute back the value of u:
Distance = (3/2)sin(2t) + C
Step 8: Evaluate the integral over the given time interval:
Distance = [(3/2)sin(2t)] between 0 and π/8
Step 9: Plug in the upper and lower limits:
Distance = (3/2)sin(2(π/8)) - (3/2)sin(2(0))
Simplifying,
Distance = (3/2)sin(π/4) - (3/2)sin(0)
Step 10: Evaluate the trigonometric functions:
Distance = (3/2)(1/√2) - (3/2)(0)
Simplifying further,
Distance = (3√2/4) - 0
Step 11: Final answer:
The distance traveled by the particle over the time interval 0 ≤ t ≤ π/8 is (3√2/4) meters.
To find the distance traveled by the particle over the given time interval, we need to integrate the absolute value of the velocity function with respect to time.
The absolute value of the velocity function, |v(t)|, can be calculated by taking the absolute value of each term within the function. In this case, |v(t)| = |3cos(2t)| = 3|cos(2t)|.
To determine the integral bounds for this question, we start with the given time interval, which is 0 ≤ t ≤ π/8. Plugging these values into the velocity function, we can find the bounds of integration for our problem.
v(0) = 3cos(2(0)) = 3cos(0) = 3(1) = 3 m/s
v(π/8) = 3cos(2(π/8)) = 3cos(π/4) = 3(√2/2) = (3√2)/2 m/s
Now, we can set up the integral for the distance traveled:
distance = ∫[0,π/8] |v(t)| dt
distance = ∫[0,π/8] 3|cos(2t)| dt
Next, let's focus on the integral part ∫ |cos(2t)| dt. Since the absolute function removes the negative part of the cosine wave, we are only interested in the positive part, which occurs between 0 ≤ cos(2t) ≤ 1/√2.
Hence, we can rewrite the integral as:
distance = 3 ∫[0,π/8] cos(2t) dt
To solve this integral, we use the identity ∫ cos(u) du = sin(u) + C, where C is the constant of integration.
Applying the substitution u = 2t, we have:
du = 2 dt
dt = du/2
Substituting this into the integral, we get:
distance = 3 ∫[0,π/8] cos(u) (du/2)
distance = (3/2) ∫[0,π/8] cos(u) du
Integrating cos(u) with respect to u gives us:
distance = (3/2) [sin(u)] [0,π/8]
distance = (3/2) [sin(2t)] [0,π/8]
Finally, substituting the upper and lower limits into the equation, we get the distance traveled:
distance = (3/2) [sin(2(π/8)) - sin(2(0))]
distance = (3/2) [sin(π/4) - sin(0)]
distance = (3/2) [(√2/2) - 0]
distance = (3√2)/4
Therefore, the distance traveled by the particle over the time interval 0 ≤ t ≤ π/8 is (3√2)/4 meters.