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?

Question

Ezra Adams · Accepted Answer

The particle changes direction for one value of t: #t=1#.

If the acceleration of the particle is #a(t) = 3t^2 - 2t#, then the velocity 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 #0# is a potential turning point: #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 #t=0# because that is when the particle starts moving): #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# to #t=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 #t=1#. Final Answer

Samantha Adkison · Answer

undefined The particle changes direction when its velocity changes sign. To find where this occurs, we need to find the roots of the velocity function, which is the integral of the acceleration function. So, integrate $a(t) = 3t^2 - 2t$ to get the velocity function $v(t)$. Then, find the roots of $v(t)$.