Old Midterm 1 - Solutions

Old Midterm 1 - Solutions - oldmidterm 01 – FRENNEA, KYLE...

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Unformatted text preview: 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 = 20 m. 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 v ( t ) time × (3 s) velocity × (6m / 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 = 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 + 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 + 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 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, 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 = ct 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 ....
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This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Old Midterm 1 - Solutions - oldmidterm 01 – FRENNEA, KYLE...

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