Question

A body of mass m moves in a horizontal direction such that at time t its position is given by x(t)=at⁴+bt³+ct, where a, b, and c are constants.

(a) What is the acceleration of the body?

a) 12at² + 6bt + 2c
b) 4at³ + 3bt² + 2ct
c) 24at + 6b
d) 4a + 3b

Answer 1

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.