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AP Calculus BC Full Practice Exam 5 Section II Part A

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AP Calculus BC Full Practice Exam 5 Section II Part A

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AP Calculus BC Full Practice Exam 5 Section II Part A

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Examine the function $$f$$ defined as $$f(x)=\\frac{x+\\sqrt{x}}{4},\\ 0\\le x\\le 8$$.

a) Use a Riemann sum with four equal subintervals evaluated at the midpoint to approximate the area under $$f$$ from $$x=0$$ to $$x=8$$.

b) Again using four equal subintervals, use the trapezoidal rule to approximate the area under $$f$$ from $$x=0$$ to $$x=8$$.

c) Using your result from part (b), approximate the average value of the function $$f$$ from $$x=0$$ to $$x=8$$.

d) Determine the actual average value of the function $$f$$ from $$x=0$$ to $$x=8$$.

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Check Answer Explanation

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Let $$v(t)$$ be the velocity, in meters per second, of a car at time $$t$$ seconds, $$t\\ge0$$. At time $$t=0$$, the car is traveling at $$30$$ meters per second. The driver applies the brakes so that the car's velocity satisfies the differential equation $$\\frac{dv}{dt}=-\\frac{1}{10}t-3$$

a) Find an expression for $$v$$ in terms of $$t$$, where $$t$$ is measured in seconds.

b) How long does it take for the car to come to a complete stop?

c) Write an equation for the tangent line to the velocity curve at $$t=5$$ seconds.

d) Find the total distance traveled from $$t=0$$ until the car stops.

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Given $$\\frac{dv}{dt}=-\\frac{1}{10}t-3$$ Integrate: $$v(t)=-\\frac{1}{20}t^2-3t+C$$ Use initial condition $$v(0)=30$$ $$C=30$$ Thus $$v(t)=-\\frac{1}{20}t^2-3t+30$$ Part b: Set velocity equal to zero: $$-\\frac{1}{20}t^2-3t+30=0$$ Multiply by $$20$$: $$-t^2-60t+600=0$$ $$t^2+60t-600=0$$ Solve: $$t=\\frac{-60\\pm\\sqrt{3600+2400}}{2}$$ $$t=\\frac{-60\\pm\\sqrt{6000}}{2}$$ $$t=\\frac{-60+20\\sqrt{15}}{2}$$ $$t=-30+10\\sqrt{15}$$ $$t\\approx8.73$$ Part c: Slope at $$t=5$$: $$v'(t)=-\\frac{1}{10}t-3$$ $$v'(5)=-\\frac{1}{2}-3=-\\frac{7}{2}$$ Velocity at $$5$$: $$v(5)=-\\frac{25}{20}-15+30$$ $$v(5)=13.75$$ Tangent line: $$v-13.75=-\\frac{7}{2}(t-5)$$ Part d: Distance traveled: $$\\int_0^{t}v(t)\\,dt$$ $$\\int\\left(-\\frac{1}{20}t^2-3t+30\\right)dt$$ $$=-\\frac{1}{60}t^3-\\frac{3}{2}t^2+30t$$ Evaluate from $$0$$ to $$-30+10\\sqrt{15}$$ Distance $$\\approx144.1$$ Final answers: a) $$v(t)=-\\frac{1}{20}t^2-3t+30$$ b) $$t=-30+10\\sqrt{15}\\approx8.73$$ c) $$v-13.75=-\\frac{7}{2}(t-5)$$ d) $$\\approx144.1$$

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a) $$v(t)=-\\frac{1}{20}t^2-3t+30$$ b) $$t=-30+10\\sqrt{15}\\approx8.73$$ c) $$v-13.75=-\\frac{7}{2}(t-5)$$ d) $$\\approx144.1$$

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Correct

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Section II Part A

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