A particle moves along a horizontal path. Its position in meters after t seconds is given by: s(t) = t 3 − 12t + 3. Answer the following questions. (a) Find the velocity function (b) Find the acceleration function (c) When is the particle moving forward? (d) When is the particle moving backwards? (e) Find the distance that the particle travels in the time interval 0 ≤ t ≤ 3.

(a) velocity function: 3t² - 12

(b) acceleration function: 6t

(c) the particle is moving forward when t > 2

(d) the particle is moving backward when t < 2

(e) the particle travels -9 m

Explanation:

position:  s(t) = t³ − 12t + 3.

(a) by definition, velocity = ds/dt = 3t² - 12

(b) by definition, acceleration = d²s/dt² = 6t

(c) A particle moves forward when its velocity is positive. That is:

3t² - 12 > 0

3t² > 12

t² > 12/3

t² > 4

t > 2

(d) A particle moves backwards when its velocity is negative. That is:

3t² - 12 < 0

3t² < 12

t² < 12/3

t² < 4

t < 2

(e) The distance is computed as follows:

-9

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

we need to know the time interval during which the force acts. impulse of force = momentum change