Speeding up & Slowing down

Thank you for your response. What about the time interval when t is smaller than zero?

Hi,

Thank you for your question.

I would say that the easiest way to determine whether the particle is speeding up or slowing down is to see whether the velocity and acceleration have the same sign or not. In other words, when velocity and acceleration have the same sign, the object is speeding up, and when velocity and acceleration have the opposite sign, the object is slowing down. Here is how I would solve this problem:

s(t) = t^3 - 9t^2

v(t) = s'(t) = 3t^2 - 18t

a(t) = v'(t) = 6t - 18

Find the x-intercepts. of v(t) and a(t) to determine their signs:

v(t) = 0

3t^2 - 18t = 0

Velocity is zero when t = 0, t = 6

a(t) = 0

6t - 18 = 0

Acceleration is zero when t = 3.

Now, test some points (one for each intervals):

i) when 0 < t < 3 (ex: t = 1):

v(1) = negative

a(1) = negative

Therefore, speeding up (they have same sign)

ii) when 3 < t < 6 (ex: t = 5):

v(5) = negative

a(5) = positive

Therefore, slowing down (they have opposite sign)

iii) when t > 6 (ex: t = 7): v(7) = positive

a(7) = positive

Therefore, speeding up (they have same sign)

So, in conclusion, the particle is speeding up when 0 < t < 3 and t > 6, and the particle is slowing down when 3 < t < 6 .

Does this answer your question?