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

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Step-by-step 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.

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