Particle Speeding Up and Slowing Down

[Jason76]

I think this has something to do with the acceleration function a(t), which is the 2nd derivative of the s(t) position function, and the 1st derivative of the v(t) velocity function:

(h) When is the particle speeding up? When is it slowing down? (Write your answers in interval notation.)

Here is the info needed to solve the problem: "A particle moves according to the law of motion s = f(t) = t^3 - 15t^2 + 72t, t > 0, where t is measured in seconds and s in feet."

 

[Ishuda]

I think this has something to do with the acceleration function a(t), which is the 2nd derivative of the s(t) position function, and the 1st derivative of the v(t) velocity function:

(h) When is the particle speeding up? When is it slowing down? (Write your answers in interval notation.)

Here is the info needed to solve the problem: "A particle moves according to the law of motion s = f(t) = t^3 - 15t^2 + 72t, t > 0, where t is measured in seconds and s in feet."

You are correct in that it has to do with a(t). When a(t) is positive, the particle is speeding up, when a(t) is negative, it is slowing down. So when is the second derivative positive and when is it negative.

 

[Jason76]

You are correct in that it has to do with a(t). When a(t) is positive, the particle is speeding up, when a(t) is negative, it is slowing down. So when is the second derivative positive and when is it negative.

$\displaystyle 6t - 30 < 0$

$\displaystyle t < 5$ it is negative

$\displaystyle 6t - 30 > 0$

$\displaystyle t > 5$ it is positive

However, this is wrong on homework.

 

[Deleted member 4993]

$\displaystyle 6t - 30 < 0$

$\displaystyle t < 5$ it is negative

$\displaystyle 6t - 30 > 0$

$\displaystyle t > 5$ it is positive

However, this is wrong on homework.

v = 3t² - 30t + 72 = 3(t-4)(t-6)

The velocity is negative between t=4 and t=6

That means in that interval the particle is traveling in the opposite direction.

What does that mean for your problem?