I need help with some Calculus problems?

Question

1. A particle moves on the x-axis so that its position is given by x(t) = t^4-6t^2+8 for t>=0. For what times(t) is the velocity of the particle increasing?

2. A particle moves on the x-axis such that its position at any time t > 0 is given by x(t) = t^3-9t^2+24t. What is the velocity of the particle when its acceleration is zero? 

 

Answer 1

2.

The position of the particle is x(t) = t^3 - 9t^2 +24t.

The velocity of the particle is v(t) = dx(t)/dt.

v(t) = dx(t) /dt = 3t^2 -18t +24

Acceleration of the particle is a(t) 

Since acceleration  = 0

6t -18 = 0

6t = 18

t= 3.

The velocity of the particle when acceleration is zero is

v(t) = 3t^2 - 18t +24 

v(t) = 3[(3)^2] -18 (3) +24[since t = 3]

v(t) = 27 - 54 +24

v(t) = 51 - 54

v(t) = - 3.

Hence, the velocity of the particle when acceleration is zero is v(t) = - 3.

Answer 2

1.

The position of the particle on x-axis is x(t) =  t^4 - 6t^ 2+ 8; for t>0

Velocity of the particle v(t) = dx(t)/dt

  dx(t)/dt = 4t^3 - 12 t

To find whether the velocity of the particle is increasing or decreasing ,dx(t)/dt = 0

4t^3 -12t=0

4t^3 = 12t

t^2 = 3