Unit 5 Quiz APCALC-061-T001: AP Calculus AB Part 1 (TL) / Unit 5 – Applications of Derivatives 3. What is the acceleration equation of a particle that has the following distance equation? x(t) = t \ln t + \frac{1}{3}t^3 \frac{1}{t} + 2t \frac{1}{t} + t^2 t^2 – \ln t – 1 t^2 + \ln t + 1 \frac{1}{t} + 2t + 1
The Correct Answer and Explanation is:
The correct answer is:
1/t + 2t
Explanation
To find the acceleration equation of a particle, we must take the second derivative of its distance (or position) equation with respect to time. The first derivative of the position function gives the velocity, and the second derivative gives the acceleration.
The given distance equation is:
x(t) = t ln t + (1/3)t³
Step 1: Find the velocity equation, v(t), by taking the first derivative of x(t).
We will differentiate the equation term by term.
For the first term, t ln t, we must use the product rule, which states that the derivative of f(t)g(t) is f'(t)g(t) + f(t)g'(t).
Let f(t) = t and g(t) = ln t.
Then f'(t) = 1 and g'(t) = 1/t.
Applying the product rule:
d/dt (t ln t) = (1)(ln t) + (t)(1/t) = ln t + 1
For the second term, (1/3)t³, we use the power rule.
d/dt ((1/3)t³) = 3 * (1/3) * t^(3-1) = 1 * t² = t²
Combining these results, the velocity equation is:
v(t) = x'(t) = (ln t + 1) + t² = ln t + t² + 1
Step 2: Find the acceleration equation, a(t), by taking the derivative of the velocity equation, v(t).
Now we differentiate v(t) = ln t + t² + 1 term by term.
The derivative of ln t is 1/t.
The derivative of t² using the power rule is 2t.
The derivative of the constant 1 is 0.
Combining these results, the acceleration equation is:
a(t) = v'(t) = x”(t) = 1/t + 2t + 0
Therefore, the final acceleration equation is:
a(t) = 1/t + 2t
