How many values of t does the particle change direction if a particle moves with acceleration a(t)=3t^2-2t and it's initial velocity is 0?
1 Answer
The particle changes direction for one value of t:
Explanation:
If the acceleration of the particle is
inta(t)dt = int(3t^2-2t)dt = t^3 - t^2 + C
Since the initial velocity is 0 (when t=0):
0^3 - 0^2 + C = 0
C = 0 So our equation simplifies to
v(t) = t^3 - t^2
Every point where the velocity is
v(t) = 0
t^3 - t^2 = 0
t^2(t-1) = 0
t = 0 " " and " " t=1
To check whether the particle changes directions at each of these points, we need to pick test points to check the intervals between them (we don't have to check before
v(color(red)(1/2)) = (color(red)(1/2))^3 - (color(red)(1/2))^2 = 1/8 - 1/4 = -1/8 So in the interval from
t=0 tot=1 , the particle moves in the negative direction.
v(color(blue)2) = (color(blue)2)^3 - (color(blue)2)^2 = 8-4 = 4 So in the interval beyond
t=1 , the particle moves in the positive direction.
Therefore, the particle changes direction at
Final Answer
