Calculus Chapter 7 Exam

Calculus Key Chapter 7 - Exam Problems Day 1: (1-6) 15, 18 Day 2: (9, 11, 16, 19, 21, 22) Position - Velocity - Acceleration Other packet Day 3: (25-26) (27-39) Day 4: (45-54) odd D 1. The velocity of a particle moving on a line at time t is (v(t) = 3t^{\frac{1}{2}} + 5t^{\frac{3}{2}}) meters per second. How Day 5: (61, 62, 64, 65, 70, 71) many meters did the particle travel from t = 0 to t = 4? Other Days: 7, 16, 56-69 A) 32 B) 40 C) 64 D) 80 E) 184 A 2. A particle with velocity at any time t given by (v(t) = e^t) moves in a straight line. How far does the particle move from t = 0 to t = 2? A) (e^2 - 1) B) e - 1 C) 2e 1 D) (e^2) E) (\frac{e^3}{3}) C 3. The acceleration of a particle moving along the x-axis at time t is given by a(t) = 6t - 2. If the velocity is 25 when t = 3 and the position is 10 when t = 1, then the position x(t) = A) (9t^2 + 1) B) (3t^2 - 2t + 4) C) (t^3 - t^2 + 4t + 6) D) (t^3 - t^2 + 9t - 20) E) (36t^3 - 4t^2 - 77t + 55) 2 C 4. If the velocity of a particle moving along the x-axis is v(t) = 2t - 4 and if at t = 0 its position is 4, then at any time t its position x(t) is 3 A) (t^2 - t) B) (t^2 - 4t - 4) C) (t^2 - 4t + 4) D) (2t^2 - 4t) E) (2t^2 - 4t + 4) B 5. A particle travels in a straight line with a constant acceleration of 3 meters per second per second. If the velocity of the particle is 10 meters per second at 4 time 2 seconds, how far (in meters) does the particle travel during the time interval when its velocity increases from 4 meters per second to 10 meters 5 per second? A) 20 B) 14 C) 7 D) 6 E) 3 E 6. A particle moves along the x-axis so that at any time (t \ge 0) the acceleration of the particle is (a(t) = e^{-2t}). If at t = 0 the velocity of the particle is (\frac{5}{2}) and its position is (\frac{17}{4}), then its position at any time t > 0 is x(t) = A) (\frac{-e^{-2t}}{2} 6 + 3) B) (\frac{e^{-2t}}{4} + 4) C) (4e^{-2t} + \frac{9}{2}t + \frac{1}{4}) D) (\frac{e^{-2t}}{4} + 3t + \frac{15}{4}) E) (\frac{e^{-2t}}{4} + 3t + 4)   (1985 BC1) A particle moves along the x-axis with acceleration given by a(t) = cos t for t ≥ 0. At t = 0, the velocity v(0) of the particle is 2 and the position x(0) is 5. (a) Write an expression for the velocity v(t) of the particle. (b) Write an expression for the position x(t). (c) For what values of t is the particle moving to the right? Justify your answer. (d) Find the total distance traveled by the particle from t = 0 to t = π/2.   Solutions: (a) v(t) = ∫ a(t) dt = ∫ cos t dt = sin t + C v(0) = 2 = sin 0 + C = C = 2 v(t) = sin t + 2 (b) x(t) = ∫ v(t) dt = ∫ (sin t + 2) dt = -cos t + 2t + C x(0) = 5 = -cos 0 + 0 + C = -1 + C = C = 6 x(t) = -cos t + 2t + 6 (c) v(t) > 0, sin t + 2 > 0. Since sin t ≥ -1, sin t + 2 ≥ 1, so v(t) is always greater than 0. The particle is always moving right. (d) ∫₀^(π/2) |v(t)| dt = ∫₀^(π/2) (sin t + 2) dt = (-cos t + 2t) |₀^(π/2) = (-cos(π/2) + π) - (-cos 0 + 0) = (0 + π) - (-1 + 0) = π + 1 (1986 BC1) A particle moves along the x-axis so that, at any time t ≥ 1, its acceleration is given by a(t) = 1/t. At time t = 1, the velocity of the particle is -2 and its position is 4. (a) Find the velocity v(t) for t ≥ 1. (b) Find the position x(t) for t ≥ 1. (c) What is the position of the particle when it is farthest to the left? Solutions: (a) v(t) = ∫ (1/t) dt = ln|t| + C v(1) = -2 = ln 1 + C = 0 + C = C = -2 v(t) = ln|t| - 2 (b) x(t) = ∫ (ln t - 2) dt = t ln t - t - 2t + C = t ln t - 3t + C x(1) = 4 = 1 ln 1 - 3 + C = 0 - 3 + C = C = 7 x(t) = t ln t - 3t + 7 (c) v(t) = 0, ln t - 2 = 0, ln t = 2, t = e² x(e²) = e² ln e² - 3e² + 7 = 2e² - 3e² + 7 = 7 - e² (1990 AB1) A particle, initially at rest, moves along the x-axis so that its acceleration at any time t ≥ 0 is given by a(t) = 12t² - 4. The position of the particle when t = 1 is x(1) = 3. (a) Find the values of t for which the particle is at rest. (b) Write an expression for the position x(t) of the particle at any time t ≥ 0. (c) Find the total distance traveled by the particle from t = 0 to t = 2.   Solutions: (a) v(t) = ∫ (12t² - 4) dt = 4t³ - 4t + C v(0) = 0 = C v(t) = 4t³ - 4t v(t) = 0, 4t(t² - 1) = 0, t = 0, t = 1, t = -1 (not in domain) (b) x(t) = ∫ (4t³ - 4t) dt = t⁴ - 2t² + C x(1) = 3 = 1 - 2 + C = C = 4 x(t) = t⁴ - 2t² + 4 (c) x(0) = 4, x(1) = 3, x(2) = 16 - 8 + 4 = 12 ∫₀² |4t³ - 4t| dt = ∫₀¹ (4t - 4t³) dt + ∫₁² (4t³ - 4t) dt = (2t² - t⁴) |₀¹ + (t⁴ - 2t²) |₁² = (2 - 1) - 0 + (16 - 8) - (1 - 2) = 1 + 8 + 1 = 10 (1991 BC1) A particle moves along the x-axis so that its velocity at any time t ≥ 0 is given by v(t) = 12t² - 36t + 15. At t = 1, the particle is at the origin. (a) Find the position x(t) of the particle at any time t ≥ 0. (b) Find all values of t for which the particle is at rest. (c) Find the maximum velocity of the particle for 0 ≤ t ≤ 2. (d) Find the total distance traveled by the particle from t = 0 to t = 2.   Solutions: (a) x(t) = ∫ (12t² - 36t + 15) dt = 4t³ - 18t² + 15t + C x(1) = 0 = 4 - 18 + 15 + C = 1 + C = C = -1 x(t) = 4t³ - 18t² + 15t - 1 (b) v(t) = 0, 12t² - 36t + 15 = 0, 4t² - 12t + 5 = 0, (2t - 1)(2t - 5) = 0, t = 1/2, t = 5/2 (not in domain) (c) v'(t) = 24t - 36 = 0, t = 36/24 = 3/2 v(0) = 15, v(1/2) = 3 - 9 + 15 = 9, v(3/2) = 27 - 81/2 + 15 = -9/4, v(2) = 48 - 72 + 15 = -9 Maximum velocity is 15. (d) ∫₀² |12t² - 36t + 15| dt = ∫₀^(1/2) (12t² - 36t + 15) dt - ∫_(1/2)^(3/2) (12t² - 36t + 15) dt + ∫_(3/2)² (12t² - 36t + 15) dt = ... (1994 AB4) A particle moves along the x-axis so that its velocity at an