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Chapter 2: Problem 189

The motion of a particle along a straight line is described by the function \(\mathrm{x}=(3 \mathrm{t}-2)^{2}\). Calculate the acceleration after $10 \mathrm{~s}$. (A) \(9 \mathrm{~ms}^{-2}\) (B) \(18 \mathrm{~ms}^{-2}\) (C) \(36 \mathrm{~ms}^{-2}\) (D) \(6 \mathrm{~ms}^{-2}\)

Short Answer

Expert verified
The acceleration after 10 seconds is (B) \(18 \mathrm{~ms}^{-2}\).

Step by step solution

01

Find the velocity function, v(t)

To find the velocity function, we take the first derivative of the position function, x(t), with respect to time (t): \(v(t) = \frac{d}{dt}(3t-2)^2\) To do this, we will use the chain rule, which states that \(\frac{d}{dt}(f(g(t))) = f'(g(t)) * g'(t)\): In this case, our function x(t) is the function f(g(t)) with an inner function of g(t) = 3t - 2, and an outer function f(u) = u^2. First, we find the derivative of the outer function: \(f'(u) = 2u\) Next, we find the derivative of the inner function: \(g'(t) = 3\) Now, applying the chain rule, we obtain the velocity function v(t): \(v(t) = f'(g(t)) * g'(t) = 2(3t - 2)(3) = 6(3t - 2)\)
02

Find the acceleration function, a(t)

To find the acceleration function, we take the first derivative of the velocity function, v(t), with respect to time (t): \(a(t) = \frac{d}{dt}(6(3t-2))\) Now by applying the power rule, we get: \(a(t) = 6(\frac{d}{dt} (3t-2))\) Notice that the term inside the derivative, (3t - 2), has a derivative of 3. So, we multiply the 6 by the 3 and get: \(a(t) = 18\)
03

Find the acceleration after 10 seconds

The acceleration function a(t) is simply equal to \(18 \mathrm{~ms}^{-2}\). Since it is a constant function, it does not depend on the time (t). Therefore, the acceleration after 10 seconds is: \(a(10) = 18 \mathrm{~ms}^{-2}\) The correct answer is (B) \(18 \mathrm{~ms}^{-2}\).

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