A particle moves along the x-axis with velocity v(t) = sin(2t), with t measured in seconds and v(t) measured in feet per second. Find the total distance travelled by the particle from t = 0 to t = π seconds.
Do I have to take the integral of the equation like ∫ sin(2t) where a=0 b=pi
whats the answer lol
you don't know the answer then dont say it you nerd
idk either guys
Yes, you are correct. To find the total distance travelled by the particle, you need to take the integral of the absolute value of the velocity function over the given time interval.
In this case, the velocity function is v(t) = sin(2t), and you want to find the total distance travelled from t = 0 to t = π.
To do this, you can set up the integral as follows:
Total distance = ∫[0 to π] |v(t)| dt
Since the velocity function is positive for the entire interval [0, π], taking the absolute value is not necessary.
Total distance = ∫[0 to π] v(t) dt
Using the given velocity function, we can now evaluate the integral:
Total distance = ∫[0 to π] sin(2t) dt
To solve this integral, you can use the substitution method. Let u = 2t, which implies du = 2 dt. Rearranging, we have dt = du/2.
Now, let's substitute these values back into the integral:
Total distance = ∫[0 to π] sin(u) (du/2)
Simplifying, we get:
Total distance = (1/2) ∫[0 to π] sin(u) du
Now, integrating the function sin(u) with respect to u gives us:
Total distance = (1/2) [-cos(u)] [0 to π]
= (1/2) [(-cos(π)) - (-cos(0))]
= (1/2) [(1) - (-1)]
= (1/2) [2]
= 1
Therefore, the total distance travelled by the particle from t = 0 to t = π seconds is 1 foot.
He specifically says it in the end, learn how to actually ready before making comments like that Pablo lol.
The answer is 0.
since v(t) = sin (2t)
then s(t) = -(1/2)cos(2t) + c , where s(t) is the distance
when t = 0,
s(0) = -(1/2)cos 0 + c
= c - 1/2
when t = π
s(π) = -(1/2)cos 2π + c
= c - 1/2
distance traveled = s(π) - s(0)
= c - 1/2 - (c - 1/2)
= 0
which is exactly what you would get.
Just a different way of writing it.