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Old Midterm 1 - Solutions

Old Midterm 1 - Solutions - oldmidterm 01 FRENNEA KYLE Due...

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oldmidterm 01 – FRENNEA, KYLE – Due: Feb 12 2007, 4:00 am 1 Question 1 part 1 of 2 10 points The velocity v ( t ) of some particle is plotted as a function of time on the graph below. The scale on the horizontal axis is 3 s per grid square and on the vertical axis 6 m / s per grid square. Initially, at t = 0 the particle is at x 0 = 20 m. 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 8 9 v ( t ) time × (3 s) velocity × (6 m / s) What is the position x of the particle at time t = 12 s? Correct answer: 200 m (tolerance ± 1 %). Explanation: Looking at the v ( t ) plot we see that over time t = 4 × 3 s = 12 s, the particle’s velocity decreases from the initial v 0 = 4 × 6 m / s = 24 m / s to final v f = 1 × 6 m / s = 6 m / s. The v ( t ) line is straight, which indicates constant deceleration rate, hence the average velocity is given by ¯ v = v 0 + v f 2 = (24 m / s) + (6 m / s) 2 = 15 m / s . Consequently, the particle’s displacement during this time is simply Δ x = t ¯ v = (12 s) (15 m / s) = 180 m , and its final position x = x 0 + Δ x = (20 m) + (180 m) = 200 m . Question 2 part 2 of 2 10 points What is the particle’s acceleration? Correct answer: 1 . 5 m / s 2 (tolerance ± 1 %). Explanation: The average acceleration of the particle is ¯ a = Δ v Δ t = v f v 0 t = (6 m / s) (24 m / s) (12 s) = 1 . 5 m / s 2 . The acceleration is the slope of the velocity vs time graph. Since the v ( t ) line is straight, the acceleration is constant, hence a = ¯ a = 1 . 5 m / s 2 . Question 3 part 1 of 1 10 points A ranger in a national park is driving at 28 mi / h when a deer jumps into the road 198 ft ahead of the vehicle. After a reaction time of t the ranger applies the brakes to produce and acceleration of 9 . 14 ft / s 2 . What is the maximum reaction time al- lowed if she is to avoid hitting the deer? Correct answer: 2 . 57489 s (tolerance ± 1 %). Explanation: Given : v i = 28 mi / h = 41 . 0667 ft / s , v f = 0 , and a = 9 . 14 ft / s 2 . The total distance covered is defined by v i t + Δ x stop = Δ x ,
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oldmidterm 01 – FRENNEA, KYLE – Due: Feb 12 2007, 4:00 am 2 where t is the reaction time. v 2 f = v 2 i + 2 a Δ x = 0 since v f = 0 mi/h, so Δ x stop = v 2 i 2 a . Thus v i t = Δ x Δ x stop t = Δ x v i Δ x stop v i = Δ x v i + v 2 i 2 a v i = Δ x v i + v i 2 a = 198 ft 41 . 0667 ft / s + 41 . 0667 ft / s 2 ( 9 . 14 ft / s 2 ) = 2 . 57489 s . Question 4 part 1 of 1 10 points The height of a helicopter above the ground is given by h = c t 3 , where c = 1 . 6 m / s 3 , h is in meters, and t is in seconds. The helicopter takes off at t = 0 s. After 2 s it releases a small mailbag. The acceleration of gravity is 9 . 8 m / s 2 . How long after its release does the mailbag reach the ground? Correct answer: 4 . 499 s (tolerance ± 1 %). Explanation: Given : t = 2 s . Under free fall, h ( t ) = y ( t ) = y 0 + v 0 t + 1 2 a t 2 . The initial height of the mailbag is the height of the helicopter 2 s after takeoff h 0 = h ( t ) = (1 . 6 m / s 3 ) (2 s) 3 = 12 . 8 m , and it starts its free fall motion from this point. Its initial velocity is equal to the veloc- ity of the helicopter at that time v 0 = dh dt = 3 c t 2 = 3 (1 . 6 m / s 3 ) (2 s) 2 = 19 . 2 m / s .
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