Kaplunovsky - final

# Kaplunovsky - final - final 01 – – Due: May 10 2007,...

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Unformatted text preview: final 01 – – Due: May 10 2007, 11:00 pm 1 Question 1, chap 1, sect 6. part 1 of 1 6 points Given 1 inch ≡ 2 . 54 cm and 1 foot ≡ 12 inches, how many square centimeters are there in 2 . 48 ft 2 ? 1. 75 . 5904 cm 2 2. 1 . 33333 cm 2 3. 2304 cm 2 correct 4. . 0813648 cm 2 5. . 00266945 cm 2 6. 192 cm 2 7. 55 . 3537 cm 2 8. . 0067804 cm 2 Explanation: 2 . 48 ft 2 × (12 inch ft ) 2 × (2 . 54 cm inch ) 2 = 2304 cm 2 Question 2, chap 2, sect 2. part 1 of 1 6 points The graph shows position as a function of time for two trains ( A and B ) running on par- allel tracks. At time t =0 the starting position of both trains are at the origin (position zero). time position t p A B Which is true: 1. Both trains speed up all the time 2. Both trains have the same velocity at some time before t p correct 3. At time t p , both trains have the same velocity 4. Somewhere before time t p , both trains have the same acceleration 5. In the time interval from t =0 to t = t p , train B covers more distance than train A Explanation: The slope of the position vs time curve indicates velocity. The slope of the curve B is parallel to line A at some point 0 < t v < t p , as shown below. time position t p A B t v Positions are identical at t p . Velocities are identical at t v . Question 3, chap 2, sect 5. part 1 of 1 8 points A ranger in a national park is driving at 56 km / h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction time of t s, the ranger applies the brakes to produce an acceleration of − 3 . 0 m/s 2 . What is the maximum reaction time al- lowed if the ranger is to avoid hitting the deer? 1. 1 . 33419 s 2. 1 . 37607 s 3. 1 . 42659 s 4. 1 . 48046 s 5. 1 . 52857 s 6. 1 . 58598 s correct 7. 1 . 64371 s 8. 1 . 71398 s final 01 – – Due: May 10 2007, 11:00 pm 2 9. 1 . 7737 s 10. 1 . 82998 s Explanation: Basic Concepts: v f = v i + a Δ t Δ x = v i Δ t + 1 2 a (Δ t ) 2 Given: v i = 56 km / h · 1 h 3600 s · 1000 m 1 km = 15 . 5556 m / s v f = 0 m / s a = − 3 . 0 m / s 2 Δ x tot = 65 m Solution: For the time interval during which the ranger decelerates, ( v f = 0 m/s), a Δ t = − v i Δ t = − v i a = − 15 . 5556 m / s − 3 m / s 2 = 5 . 18519 s . The distance he travels during the decelera- tion is Δ x = v i Δ t + 1 2 a (Δ t ) 2 = (15 . 5556 m / s) (5 . 18519 s) + ( − 3 m / s 2 ) (5 . 18519 s) 2 2 = 40 . 3292 m . The maximum reaction distance is Δ x max = Δ x tot − Δ x = 65 m − 40 . 3292 m = 24 . 6708 m . and the maximum reaction time is Δ t max = Δ x max v i = 24 . 6708 m 15 . 5556 m / s = 1 . 58598 s . Question 4, chap 4, sect 4. part 1 of 1 10 points To win the game, a place kicker must kick a football from a point 19 m (20 . 7784 yd) from the goal, and the ball must clear the crossbar, which is 3 . 05 m high. When kicked, the ball leaves the ground with a speed of 15 m / s at an angle of 53 ◦ from the horizontal....
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## This note was uploaded on 01/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Kaplunovsky - final - final 01 – – Due: May 10 2007,...

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