How many values of t does the particle change direction if a particle moves with acceleration a(t)=3t^2-2ta(t)=3t2−2t 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∫a(t)dt=∫(3t2−2t)dt=t3−t2+C
Since the initial velocity is 0 (when t=0):
0^3 - 0^2 + C = 003−02+C=0
C = 0C=0 So our equation simplifies to
v(t) = t^3 - t^2v(t)=t3−t2
Every point where the velocity is
v(t) = 0v(t)=0
t^3 - t^2 = 0t3−t2=0
t^2(t-1) = 0t2(t−1)=0
t = 0 " " and " " t=1t=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 = -1/8v(12)=(12)3−(12)2=18−14=−18 So in the interval from
t=0t=0 tot=1t=1 , the particle moves in the negative direction.
v(color(blue)2) = (color(blue)2)^3 - (color(blue)2)^2 = 8-4 = 4v(2)=(2)3−(2)2=8−4=4 So in the interval beyond
t=1t=1 , the particle moves in the positive direction.
Therefore, the particle changes direction at
Final Answer
