A particle moves on a verticle line so tht its coordinate at time t is y = t^3 - 12t + 3, t > or = to 0?

Question

a. Find the velocity function for this particle.
b. At what time is the particle moving upward?
c. Find the acceleration function for this particle.
When is the particle slowing down (decelerating)?

Answer 1

a. Simply differentiate with respect to time t:

dy / dt = 3 t² -12.

b. The particle moves upwards when dy / dt > 0, ie. when

3 t² -12 = 3 (t + 2) (t - 2) > 0 and t ≥ 0;

this occurs for t > 2.

c. Take the second derivative:

d²y / dt² = 6 t.

The particle is decelerating when d²y / dt² < 0, ie. for

6 t < 0.

Since t ≥ 0, the particle never decelerates.

Answer 2

The velocity is the derivative
V= 3t^2-12.
b) V must be positive 3t^2>12 so t> 2 as t>0
The acceleration is the derivative of V
@=6t
From t=0 to t=2 the velocity in absolute value decreases from 12 to 0

Answer 3

a. v = dy/dt = 3t^2 - 12

b. When v > 0 i.e.

3t^2 > 12
t^2 > 4
t > 2

c. a = dv/dt = 6t

d. When a < 0
i.e. when 6t < 0
Which never happens, so it never decelerates

Answer 4

i havent done this stuff since college
sorry.