For an object whose velocity in ft/sec is given by v(t)=2sin(t),what is its distance,in feet, travelled on the interval t=1 to t=5?
To find the distance traveled by the object on the interval t = 1 to t = 5, we need to integrate the absolute value of its velocity function over that interval.
1. Start by finding the antiderivative (integral) of the function v(t) = 2sin(t) with respect to t.
∫ v(t) dt = -2cos(t) + C
2. To determine the value of the constant C, we can use the initial condition at t = 1. Since we want to find the distance traveled on the interval t = 1 to t = 5, we need to find the position at t=1 and subtract it from the position at t=5 to obtain the displacement. The position function is the antiderivative of the velocity function.
Let's calculate the position by substituting the value of t = 1 into the antiderivative:
P(1) = -2cos(1) + C
3. Since the velocity function given is in ft/sec, the position function will be in feet. To find the distance traveled, we need to consider the absolute value of the displacement between t=1 and t=5, which is given by the difference in the positions.
Distance = |P(5) - P(1)|
Now, we have the general steps. Let's proceed to calculate the specific distances.
4. Calculate the position at t = 1 by substituting it into the antiderivative:
P(1) = -2cos(1) + C
5. Calculate the position at t = 5 by substituting it into the antiderivative:
P(5) = -2cos(5) + C
6. Find the absolute value of the difference in positions:
Distance = |P(5) - P(1)| = |-2cos(5) + C - (-2cos(1) + C)| = |-2cos(5) + 2cos(1)|
After substituting the values of t=5 and t=1 into the antiderivative and taking the absolute value, you will get the distance traveled by the object on the interval t=1 to t=5 in feet.
To find the distance traveled by the object on the interval t = 1 to t = 5, we need to integrate the absolute value of the velocity function over that interval.
Given that the velocity function is v(t) = 2sin(t), we can integrate it from t = 1 to t = 5.
∫[1 to 5] |v(t)| dt = ∫[1 to 5] |2sin(t)| dt
Since the absolute value of sin(t) is equal to |sin(t)| for all values of t, we can simplify the integral as follows:
∫[1 to 5] |v(t)| dt = ∫[1 to 5] |2sin(t)| dt = ∫[1 to 5] 2|sin(t)| dt
Now, we know that the absolute value of sin(t) is equal to sin(t) for t values between 0 and π. Therefore, we can rewrite the integral as follows:
∫[1 to 5] 2|sin(t)| dt = 2∫[1 to π] sin(t) dt
Using the integral formula for sin(t), we get:
2∫[1 to π] sin(t) dt = -2cos(t) [1 to π]
Evaluating the integral from t = 1 to t = π:
-2cos(π) - (-2cos(1))
Since cos(π) = -1 and cos(1) is a constant, this becomes:
2 - (-2cos(1)) = 2 + 2cos(1)
Therefore, the object traveled a distance of 2 + 2cos(1) feet on the interval t = 1 to t = 5.
do the integral of 2sin(t) from 1 to 5
your answer is .51 ft.