Final answer:
The acceleration of a body with position x(t) = at⁴ + bt³ + ct is a(t) = 12at² + 6bt, which is found by differentiating the position function twice. Therefore, the correct answer is a) 12at² + 6bt + 2c.
Step-by-step explanation:
The acceleration of a body whose position is given by the equation x(t) = at⁴ + bt³ + ct can be found by taking the second derivative of the position function with respect to time.
The first derivative, which is the velocity, would be v(t) = 4at³ + 3bt² + c. Taking the second derivative of this velocity function will give us the acceleration, which will be a(t) = 12at² + 6bt. The constant term c disappears upon differentiation as its derivative is zero.
The acceleration of the body can be found by taking the second derivative of the position function x(t). Given x(t) = at⁴ + bt³ + ct, the acceleration is the derivative of the velocity function, which is the derivative of x(t). So, taking the derivative of x(t) with respect to time, we get:
a(t) = 12at² + 6bt + 2c
Thus, the correct answer for the acceleration of the body is (a) 12at² + 6bt.