L02_hw - Lecture 2 Homework Problems P1. What must be your...

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Unformatted text preview: Lecture 2 Homework Problems P1. What must be your car's average speed in order to travel 235 km in 3.25 h? Solution: km h km/h cm during the time from P5. A rolling ball moves from cm to s to s. What is its average velocity? Solution: m/s P9. A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (b) the average velocity, in m/s Solution: mi 12.5 min km s s 400 m s m m m/s m s m/s P11. Two locomotives approach each other on parallel tracks. Each has a speed of 95 km/h with respect to the ground. If they are initially 8.5 km apart, how long will it be before they reach each other? Solution: 95 km/h 8.5 km m s m/s m m ! " ! Each locomotive clears distance 26.39 m/s The distance between the locomotives decreases ! ! s min min P17. A sprinter accelerates from rest to 10.0 m/s in 1.35 s. What is her acceleration (a) in m/s , and (b) in km/h Solution: m s m/s s m/s km h m/s km/h km/h P19. A sports car moving at constant speed travels 110 m in 5.0 s. If it then brakes and comes to a stop in 4.0 s, what is its acceleration in m/s ? Express the answer in terms of "g's", where 1.00g m/s Solution: # m/s s $ m/s s m/s m/s 9.8 m/s % 1 g 5.5 m/s % ? g 2 -9.8 m/ s 1g 2 -5.5 m/ s xg ! "g's" Note: We can ignore the signs here, because we are interested only in the absolute value: the acceleration expressed as a fraction of g. P21. A car accelerates from 13 m/s to 25 m/s in 6.0 s. What was its acceleration? How far did it travel in this time? Assume constant acceleration. Solution: m/s m/s s m/s m/s m/s m/s ! m P25. A car slows down uniformly from a speed of 21.0 m/s to rest in 6.00s. How far did it travel in that time? Solution: "Uniformly" means that the acceleration is constant. 0.0 m/s 21.0 m/s m/s s ! m/s m/s m/s m P33. A stone is dropped from the top of a cliff. It hits the ground below after 3.25 s. How high is the lciff? Solution: Choose: 1. Ground is where the position is zero; 2. "Things" pointing up are positive ! m/s (i-initial) m (the stone is dropped finally on the ground) m/s (negative, because it points "down") s (time it takes to fall) ? (what is the initial height) ! " " ! m/s m/s " m/s s s m/s s m 2 P37. A ballplayer catches a ball 3.0 s after throwing it vertically upward. With what speed did he throw it, and what height did it reach? Solution: Choose: 1. "UP" is "positive" m/s (the final and the initial position of the ball are the same) & ! 'the displacement is zero) s (the time of the flight) ? " ! m ! ! " s s= m/s m/s ! m/s " s s s 14.7 m/s (P45) A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.2 s later. If the speed of sound is 340 m/s, how high is the cliff? Solution: First the rock must reach the sea, then we must wait for the sound to travel back to us from the surface where the rock hit the water. The time 3.2 s includes both times - the time to drop and the time for sound to travel. Choose: 1. The sea level is where the position is zero; 2. "up" is positive Given: m/s m # m/s (sea level) is the height of the cliff, we want to find it out the time it takes the rock to fall is calculated from the following formula: " " using the given information, the equation becomes: m #" m/s m/s '() the sound travels at constant speed. so the distance the sound travels can be calculated from the formula: * * * * " * the sound travels from the sea level to the top of the cliff ! # (the final * destination), and m (the sea level) ! * # m " m/s we don't know the individual times for the rock to fall and the sound to come back, but we know that they add up to 3.2 seconds. ! * s Thus the above equation becomes: # m/s s (( Combined the two equations ( and (( allow us to determine the time it takes the rock to fall and the height of the cliff - we have to solve a set of 2 algebraic equations with two unknowns. m # # m/s m/s s '() (( Expressing from (( and substiting it into ( : # m/s s m/s m # # ! s 340 m/s m/s substituing in ( # m # m/s s m/s m m m # # # m/s m/s m s s s s " " m s s m/s s # m/s + # m/s m s # m/s s # m/s m s # m s # m/s Check the units! All seconds cancel out! m # m" # m # m # " #" m m Notice the first coefficient has a dimension of m , it is because it multiplies # , which has a dimension of m . Solving the quadratic equation: # 1 + , 1 1 , #+ , #+ ! # m # m check what the solution for the time is: # . s m/s ! t s s Obviously the time cannot be negative, thus the second value for the height, # m, makes no physical sense. Thus the answer is : m P39. A helicopter is ascending vertically with a speed of 5.20 m/s. At a height of 125 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? Given: Assuming "up" is positive. s " m/s $ " m m/s $ * m Find: t ? or in other words: Solution: " y y " " " , + ? " " m" " , m/s " m/s s Answer: s s P42. A stone is thrown vertically upward with a speed of 18.0 m/s (a) How fast is it moving when it reaches a height of 11.0 m? (b) How long is required to reach this height? (c) Why are there two answers to (b) Solution: Choose: 1. "up" is positive direction" m/s m/s (a) m ? The initial position is m m/s m m ! , m/s The stone is moving at a speed 10.4 m/s. The velocity, however, has a direction, thus there are two possible values, corresponding to the two times when the stone is 11.0 m above - once when it is going up, and second, when it is coming down. m ? " The time that corresponds to the first time when the stone is 11.0 m is: " m/s m/s 0.78 s m/s The time that corresponds to the second time when the stone is 11.0 m high is: m/s m/s s m/s P51. In Fig. 2-28 (a) during what time periods, if any, is the velocity constant? (b) At what time is the velocity greatest? (c) At what time, if any, is the velocity zero? (d) Does the object move in one direction or in both directions during the time shown? Solution: (a) The velocity is the slope of the graph when we plot vs. . In order to have a constant velocity, we must have a constant slope. The figure shows a constant slope (a straight line) from s to . s. (b) The greatest velocity is when the slope is the steepest. On the graph, it occurs at . s and . s. The slope has a different direction, which means that the velocity points in a different direction. (c) The velocity is zero, when the slope is zero, ! . s (d) Initially, the slope is positive, which corresponds to a postive velocity. After s, the slope is negative, which means that the velocity is negative. Now, we do not know which direction is positive and which is chosen to be positive, but we know that at different times the velocity pointed in opposite directions ! the object moves in both directions. (b) ...
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This document was uploaded on 09/21/2009.

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