Drill 1 · Math · Rates of Change and Motion
AP Calculus AB: Rates of Change and Motion (Drill 1) is a Math practice drill covering Rates of Change and Motion. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice interpreting derivatives as rates of change in context and analyzing the motion of a particle along a line using position, velocity, and acceleration. These AP Calculus AB skills appear on both the multiple-choice and free-response sections.
Question 1. A company's revenue, in thousands of dollars, is modeled by \( R(t) \), where \( t \) is the number of years since 2010. If \( R'(5) = 12 \), which of the following best interprets this value?
Explanation: Choice B is correct. \( R'(5) = 12 \) is the instantaneous rate of change of revenue at \( t = 5 \) (year 2015). Since \( R(t) \) is measured in thousands of dollars and \( t \) in years, \( R'(t) \) has units of thousands of dollars per year, so \( R'(5) = 12 \) means revenue is increasing at $12,000 per year in 2015. Choice A is incorrect because \( R'(5) \) is a rate of change, not the revenue value itself, that would be \( R(5) \). Choice C is incorrect because total change over an interval comes from \( R(5) - R(0) \) or the integral of \( R'(t) \), not from \( R'(5) \) alone. Choice D is incorrect because \( R'(5) \) gives no information about when a specific revenue level is reached.
Question 2. The function \( g \) is continuous on \( (-\infty, \infty) \). The sign of \( g'(x) \) is positive for \( x 2 \). Which of the following must be true?
Explanation: Choice C is correct. The particle is at rest when velocity \( v(t) = s'(t) = 0 \). Differentiating: \( v(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3) \). Setting equal to zero gives \( t = 1 \) and \( t = 3 \). Choice A is incorrect because \( v(0) = 9 \neq 0 \); the particle is moving at \( t = 0 \). Choice B is incorrect because it omits \( t = 1 \); a student who only partially factors or stops after finding one root makes this error. Choice D is incorrect and results from solving \( s(t) = 0 \), finding where the particle is at the origin, rather than \( v(t) = 0 \), which is where the particle is at rest.
Question 3. A particle moves along the \( x \)-axis with velocity \( v(t) = t^2 - 4t + 3 \) for \( t \geq 0 \). On which of the following intervals is the particle moving in the negative direction?
Explanation: Choice B is correct. The particle moves in the negative direction when \( v(t) < 0 \). Factoring: \( v(t) = (t-1)(t-3) \). This product is negative when exactly one factor is negative, which occurs on \( 1 < t < 3 \). Choice A is wrong because on \( 0 < t 0 \), the particle moves in the positive direction. Choice C doesn't work because on \( t > 3 \), both factors are positive, so \( v(t) > 0 \). Choice D falls short because the velocity roots are \( t = 1 \) and \( t = 3 \), and \( v < 0 \) only between them, on \( 0 < t < 1 \) the particle is already moving in the positive direction, so the full interval \( 0 < t < 3 \) is not correct.
Question 4. A particle moves along the \( x \)-axis with velocity \( v(t) = 3t^2 - 12t + 7 \) for \( t \geq 0 \). At \( t = 5 \), which of the following best describes the particle's motion?
Explanation: Choice A is correct. At \( t = 5 \): \( v(5) = 3(25) - 12(5) + 7 = 75 - 60 + 7 = 22 > 0 \), so the particle moves in the positive direction. Acceleration: \( a(t) = v'(t) = 6t - 12 \), so \( a(5) = 30 - 12 = 18 > 0 \). Since \( v(5) \) and \( a(5) \) are both positive (same sign), the particle is speeding up. Choice B is incorrect because \( v(5) = 22 > 0 \), not negative. Choice C is incorrect because slowing down requires \( v \) and \( a \) to have opposite signs, but both are positive here. Choice D is incorrect on both counts, the particle is moving in the positive direction and speeding up.
Question 5. A particle moves along the \( x \)-axis with velocity \( v(t) = t^2 - 6t + 8 \) on the interval \( 0 \leq t \leq 5 \). Which of the following statements is true?
Explanation: Choice C is correct. Factor: \( v(t) = (t-2)(t-4) \). The velocity changes sign at \( t = 2 \) and \( t = 4 \), both of which lie in \( [0, 5] \), so the particle changes direction twice. On \( [0,2] \): \( v > 0 \); on \( [2,4] \): \( v 0 \). Because the particle reverses direction, total distance \( = \int_0^5 |v(t)|\,dt \) is strictly greater than \( \left|\int_0^5 v(t)\,dt\right| \), since the negative portion on \( [2,4] \) cancels some positive displacement but adds to total distance. Displacement: \( \int_0^5 (t^2-6t+8)\,dt = \left[\dfrac{t^3}{3} - 3t^2 + 8t\right]_0^5 = \dfrac{125}{3} - 75 + 40 = \dfrac{20}{3} > 0 \). Choice A doesn't work because displacement equals total distance only when velocity never changes sign on the interval. Choice B falls short because the roots \( t = 2 \) and \( t = 4 \) are both in \( [0,5] \), producing two direction changes, not one. Choice D misses the mark: the displacement is \( \dfrac{20}{3} > 0 \), not negative.