The velocity of a particle tralling along the x-axis is given by v(t)=(e^-t)(sin(t)). Find the following

Question

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]

Answer 1

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

Answer 2

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.