Aparticle moves along the x-axis so that at time t its position is given by s(t) = (t + 1)(t − 3)3, t > 0. for what values of t is the velocity of the particle increasing? a. 0 < t < 1b. t > 1c. t > 0d. the velocity is never decreasing.

d. the velocity is never decreasing.

Therefore the velocity of the particle keeps increasing at a constant rate.

Since the velocity function is a Linear equation with a positive slope, the velocity is never decreasing.

Explanation:

Given;

The position function s(t) is given as;

s(t) = (t+1)(t-3)3 = 3t^2 -6t -9

Velocity = change in position/change in time = ds/dt

ds/dt = differentiating the position function

ds/dt = 6t - 6

v(t) = ds/dt = 6t -6

At t=0

v(t) = -6

At t=1

v(t) = 0

At t=2

v(t) = 6

Acceleration = dv/dt = 6

Therefore the velocity of the particle keeps increasing at a constant acceleration of 6.

Since the velocity function is a Linear equation with a positive slope, the velocity is never decreasing.

12000j is work done

unbalanced force

explanation: