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The velocity of a particle moving along the x-axis is given as a function of time by the expression v(t) = 3.0t² - 2.0t+4.0, where v is in meters per second and t is in seconds. What is the acceleration of the particle at t = 2.0 s? (A) 4.0 m/s² (B) 6.0 m/s² (C) 8.0 m/s2 (D) 10.0 m/s² (E) 12.0 m/s²"

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To determine the acceleration of the particle at the time t = 2.0 s, we must first understand that acceleration is the derivative of velocity with respect to time.

Given the velocity function v(t) = 3.0t² - 2.0t + 4.0, we can find acceleration by taking its derivative. This is based on the fundamental understanding in calculus that the rate of change of a function is given by its derivative.

The derivative of 3.0t² - 2.0t with respect to t is 6.0t - 2.0. By the chain rule, the derivative of the constant 4.0 is zero. Hence the acceleration a(t) is given by: a(t) = 6.0t - 2.0.

Now, to find the acceleration at t=2.0s, we replace t in the acceleration function a(t) with 2.0. Hence we get: a(2.0) = 6.0*2.0 - 2.0 = 10.0 m/s².

Therefore, the acceleration of the particle at t = 2.0 s is 10.0 m/s². So, the correct option is (D) 10.0 m/s².

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