The velocity of a particle tralling along the x-axis is given by v(t)=(e^-t)(sin(t)). Find the following
a. Compute the net change in distance from time t=1 to time t=3 (this is equivalent to s(3)-s(1)
b. compute the total distance traveeled from time t=0 to time t=6pi
c. Find the average velocity over the interval t E [0,6pi]
s=int v(t) dt do the integral.
b. Now net change: you have some positive displacement, and negative displcement.
You can
1) find where the velocity changes sign, then integrate those portions separately (negative distance), then change the sign, and add abs values.
2) square velocity. Integrate, then take the square root of the distance you get.
c. avg velocity is ans from a divided by 6
integral e^ax sin bx dx = (e^ax/(a^2+b^2))[a sin bx - b cos bx]
so a = -1, b = 1
(e^-t/2)(-sint-cost)
= (-1/2)e-t (sin t + cos t) + c
at t = 3
(-1/2)e^-3 (sin 3 + cos 3)
at t = 1
(-1/2)e^-1 (sin 1 + cos 1)
subtract
b)from 0 to 6 pi
sin 0 = 0 cos 0 = 1
sin 6i = 0 cos 6pi = 1
so it does not displace BUT IT MOVES (back and forth)
do it from 0 to pi/2 then from pi/2 to pi etc Use absolute value of distance, negative counts the same as positive for distance, a scalar, but not for position or "displacement" which are vectors.
c) not clear
The average vector VELOCITY is small, because it ends up about where it started. subtract the final position from the original and divide by 6 pi
However the average scalar SPEED is the distance covered from t = 0 to t = pi/2 divided by pi/2 etc because negative speed counts as much as positive.
To find the answers to these questions, we need to integrate the given velocity function v(t) to obtain the position function s(t).
a. Net Change in Distance:
First, let's find the position function s(t) by integrating v(t).
Integrating v(t) = (e^(-t))(sin(t)), we get:
s(t) = ∫(e^(-t))(sin(t)) dt
Next, find the antiderivative of v(t) to obtain s(t):
s(t) = ∫(e^(-t))(sin(t)) dt = -e^(-t)cos(t) + ∫(e^(-t))(cos(t)) dt
Now, evaluate the definite integral from t = 1 to t = 3 to find the net change in distance:
Net Change in Distance = s(3) - s(1) = [-e^(-3)cos(3) + ∫(e^(-3))(cos(3)) dt] - [-e^(-1)cos(1) + ∫(e^(-1))(cos(1)) dt]
Solving this expression will give you the net change in distance from time t = 1 to time t = 3.
b. Total Distance Traveled:
To find the total distance traveled, we need to consider both positive and negative displacements over the given time period.
The displacement from t = 0 to t = 6π is given by s(6π) - s(0).
However, since we want to find the total distance traveled, we need to consider the absolute value of the displacement.
Therefore, the total distance traveled from t = 0 to t = 6π is |s(6π) - s(0)|.
Evaluate the integral from t = 0 to t = 6π and take the absolute value of the resulting expression to find the total distance traveled.
c. Average Velocity:
The average velocity over a specific time interval is given by the total displacement divided by the duration of the interval.
To find the average velocity over the interval t ∈ [0, 6π], we need to calculate the total displacement during this time period and divide it by the duration, which is (6π - 0).
The average velocity over the interval is given by:
Average Velocity = (s(6π) - s(0)) / (6π - 0)
Evaluate the definite integral from t = 0 to t = 6π and divide the resulting expression by 6π to find the average velocity.