Two particles are moving with positions given by x= 4t^2−2t and x= 6t^3+8t respectively.

Explanation:

a)Which of the two has uniform acceleration?

Acceleration is the second derivative of position.  The acceleration of the first particle is:

x = 4t² − 2t

v = 8t − 2

a = 8

The acceleration of the second particle is:

x = 6t³ + 8t

v = 18t² + 8

a = 36t

The first particle has uniform acceleration.

b)Which one is likely to come to rest at some time during its motion?

The particles come to rest when v = 0.  The first particle's velocity has a real zero at t = 4.  The second particle's velocity has only imaginary zeros, meaning v is never 0.

im sorry i dont know

a is the best answer your welcome