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AP Calculus BC 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration - FRQs - Exam Style Questions

AP Calculus BC - FRQs and Answer - All Topics

No-Calc Question

t
(minutes)		012202440
v(t)
(meters per minute)		0200240-220150
Johanna jogs along a straight path. For \(0 \le t \le 40\), Johanna’s velocity is given by a differentiable function \(v\). Selected values of \(v(t)\), where \(t\) is measured in minutes and \(v(t)\) is measured in meters per minute, are given in the table above.
(a) Use the data in the table to estimate the value of \(v'(16)\).
(b) Using correct units, explain the meaning of the definite integral \(\displaystyle \int_{0}^{40}\!|v(t)|\,dt\) in the context of the problem.
Approximate the value of \(\displaystyle \int_{0}^{40}\!|v(t)|\,dt\) using a right Riemann sum with the four subintervals indicated in the table.
(c) Bob is riding his bicycle along the same path. For \(0 \le t \le 10\), Bob’s velocity is modeled by \(B(t)=t^{3}-6t^{2}+300\), where \(t\) is measured in minutes and \(B(t)\) is measured in meters per minute.
Find Bob’s acceleration at time \(t=5\).
(d) Based on the model \(B\) from part (c), find Bob’s average velocity during the interval \(0 \le t \le 10\).

Most-appropriate topic codes (CED):

• TOPIC 2.3: Estimating Derivatives of a Function at a Point — part (a)
• TOPIC 6.3: Riemann Sums, Summation Notation, and Definite Integral Notation — part (b)
• TOPIC 4.2: Straight-Line Motion: Connecting Position, Velocity, and Acceleration — part (c)
• TOPIC 8.1: Finding the Average Value of a Function on an Interval — part (d)

▶️ Answer/Explanation

(a)
Symmetric difference about \(t=16\) using \(t=12\) and \(t=20\):
\(\displaystyle v'(16)\approx \frac{v(20)-v(12)}{20-12}=\frac{240-200}{8}= \boxed{5\ \text{m/min}^2}.\)

(b)
Meaning: \(\displaystyle \int_{0}^{40}\!|v(t)|\,dt\) is the total distance Johanna travels (meters) for \(0\le t\le 40\).
Right Riemann sum on \([0,12],[12,20],[20,24],[24,40]\):
\(\displaystyle \int_{0}^{40}\!|v(t)|\,dt \approx 12\times|v(12)|+8\times|v(20)|+4\times|v(24)|+16\times|v(40)|\)
\(=12\times200+8\times240+4\times220+16\times150\)
\(=2400+1920+880+2400=\boxed{7600\ \text{meters}}.\)

(c)
\(B(t)=t^{3}-6t^{2}+300\ \Rightarrow\ B'(t)=3t^{2}-12t\) (acceleration, m/min\(^2\)).
\(B'(5)=3\cdot25-12\cdot5=75-60=\boxed{15\ \text{m/min}^{2}}.\)

(d)
Average velocity on \([0,10]\): \(\displaystyle \bar v=\frac{1}{10}\int_{0}^{10}B(t)\,dt\).
\(\displaystyle \int B(t)\,dt=\frac{t^{4}}{4}-2t^{3}+300t\). Evaluate \(0\to 10\):
\(\displaystyle \bar v=\frac{1}{10}\Big[\frac{10^{4}}{4}-2\cdot10^{3}+300\cdot10\Big]\) \(=\frac{1}{10}(2500-2000+3000)=\boxed{350\ \text{m/min}}.\)

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