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2022 AP Calculus AB FRQ Question 1: Toll Plaza Arrival Rate

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2022.

Question 1

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Toll Plaza Arrival Rate

Rate, Average Value & Optimization

Hard
From 5 A.M. to 10 A.M., the rate at which vehicles arrive at a certain toll plaza is given by A(t)=450sin⁡(0.62t)A(t) = 450\sqrt{\sin(0.62t)}, where tt is the number of hours after 5 A.M. and A(t)A(t) is measured in vehicles per hour. Traffic is flowing smoothly at 5 A.M. with no vehicles waiting in line.
Part AEasy1 point
Write, but do not evaluate, an integral expression that gives the total number of vehicles that arrive at the toll plaza from 6 A.M. (t=1t=1) to 10 A.M. (t=5t=5).

Answer

∫15A(t) dt=∫15450sin⁡(0.62t) dt\displaystyle\int_1^5 A(t)\,dt = \int_1^5 450\sqrt{\sin(0.62t)}\,dt
Full Solution & Work

Set up the accumulation integral

Total vehicles = accumulated arrival rate over the interval:
∫15A(t) dt=∫15450sin⁡(0.62t) dt\int_1^5 A(t)\,dt = \int_1^5 450\sqrt{\sin(0.62t)}\,dt

AP Scoring — 1 Point

**P1**: Correct integrand A(t)A(t) with correct limits t=1t=1 to t=5t=5.
Part BMedium2 points
Find the average value of the rate, in vehicles per hour, at which vehicles arrive at the toll plaza from 6 A.M. (t=1t=1) to 10 A.M. (t=5t=5).

Answer

≈ 375.537 vehicles per hour
Full Solution & Work

Apply the average value formula

15−1∫15A(t) dt\frac{1}{5-1}\int_1^5 A(t)\,dt

Evaluate with a calculator

≈375.537 vehicles per hour\approx 375.537 \text{ vehicles per hour}

AP Scoring — 2 Points

**P1**: Correct average-value setup 14∫15A(t) dt\frac14\int_1^5 A(t)\,dt.
**P2**: Correct answer, 375.537 (accept 375.536–375.538).
Part CMedium2 points
Is the rate at which vehicles arrive at the toll plaza at 6 A.M. (t=1t=1) increasing or decreasing? Give a reason for your answer.

Answer

Increasing, because A′(1)≈148.947>0A'(1)\approx148.947>0.
Full Solution & Work

Differentiate A(t)

A′(t)=450⋅12(sin⁡(0.62t))−1/2⋅0.62cos⁡(0.62t)=139.5cos⁡(0.62t)sin⁡(0.62t)A'(t) = 450\cdot\frac12\big(\sin(0.62t)\big)^{-1/2}\cdot0.62\cos(0.62t) = \frac{139.5\cos(0.62t)}{\sqrt{\sin(0.62t)}}

Evaluate at t = 1

A′(1)≈148.947>0A'(1) \approx 148.947 > 0

Conclude

Since A′(1)>0A'(1)>0, the rate at which vehicles arrive is **increasing** at t=1t=1.

AP Scoring — 2 Points

**P1**: A correct expression or numerical value for A′(1)A'(1).
**P2**: Correct conclusion ("increasing") tied to the sign of A′(1)A'(1).
Part DHard4 points
A line forms whenever A(t)≥400A(t)\geq400. The number of vehicles in line at time tt, for a≤t≤4a\leq t\leq4, is given by N(t)=∫at(A(x)−400) dxN(t)=\displaystyle\int_a^t (A(x)-400)\,dx, where aa is the time when a line first begins to form. To the nearest whole number, find the greatest number of vehicles in line at the toll plaza in the time interval a≤t≤4a\leq t\leq4. Justify your answer.

Answer

≈ 71 vehicles
Full Solution & Work

Find a, the time the line first forms

aa is the first positive solution to A(t)=400A(t)=400:
450sin⁡(0.62t)=400  ⟹  t=a≈1.469450\sqrt{\sin(0.62t)}=400 \implies t = a \approx 1.469

Find where N could be maximized

N′(t)=A(t)−400N'(t)=A(t)-400. Since AA rises above 400 at t=at=a and later falls back below 400 at a second crossing, t≈3.598t\approx3.598 (within [a,4][a,4]), N′(t)>0N'(t)>0 on (a,3.598)(a,3.598) and N′(t)<0N'(t)<0 on (3.598,4)(3.598,4). So NN is maximized at t≈3.598t\approx3.598.

Evaluate N at the maximizing time

N(3.598)=∫a3.598(A(x)−400) dx≈71.254N(3.598) = \int_a^{3.598}\big(A(x)-400\big)\,dx \approx 71.254

(compare to N(4)≈62.338N(4)\approx62.338, confirming the interior critical point gives the larger value)

Conclude

To the nearest whole number, the greatest number of vehicles in line is **71**.

AP Scoring — 4 Points

**P1**: Correctly finds a≈1.469a\approx1.469, the first time A(t)=400A(t)=400.
**P2**: Considers N′(t)=A(t)−400=0N'(t)=A(t)-400=0, i.e. the second crossing t≈3.598t\approx3.598, as the critical point of interest.
**P3**: A justification comparing NN at the critical point to N(4)N(4) (global argument), or a sign analysis of N′N' around t≈3.598t\approx3.598.
**P4**: Correct final answer, 71 vehicles, with supporting work.

Common mistake: A common shortcut error: forgetting that the domain is restricted to $[a,4]$ and instead searching for a maximum over the full $[0,5]$ range covered by $A(t)$ — the question only asks about the line's length up to $t=4$.

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