Chapter 12

# Chapter 12 - Engineering Mechanics Dynamics Chapter 12...

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Engineering Mechanics - Dynamics Chapter 12 Problem 12-1 A truck traveling along a straight road at speed v 1 , increases its speed to v 2 in time t . If its acceleration is constant, determine the distance traveled. Given: v 1 20 km hr v 2 120 km hr t 15 s Solution: a v 2 v 1 0 t a 1.852 m s 2 d v 1 t 1 2 at 2 . d 291.67 m Problem 12-2 A car starts from rest and reaches a speed v after traveling a distance d along a straight road. Determine its constant acceleration and the time of travel. Given: v 80 ft s d 500 ft Solution: v 2 2 ad a v 2 2 d a 6.4 ft s 2 v t v a t 12.5 s Problem 12-3 A baseball is thrown downward from a tower of height h with an initial speed v 0 . Determine the speed at which it hits the ground and the time of travel. Given: h 50 ft g 32.2 ft s 2 v 0 18 ft s Solution: v v 0 2 2 gh . v 59.5 ft s 1

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Engineering Mechanics - Dynamics Chapter 12 t v v 0 0 g t 1.29 s * Problem 12–4 Starting from rest, a particle moving in a straight line has an acceleration of a = ( bt + c ). What is the particle’s velocity at t 1 and what is its position at t 2 ? Given: b 2 m s 3 c 6 0 m s 2 t 1 6 s t 2 11 s Solution: a t ( ) bt c . v t ( ) 0 t t a t ( ) ´ µ d d t ( ) 0 t t v t ( ) ´ µ d v t 1 + , 0 m s d t 2 + , 80.7 m Problem 12-5 Traveling with an initial speed v 0 a car accelerates at rate a along a straight road. How long will it take to reach a speed v f ? Also, through what distance does the car travel during this time? Given: v 0 70 km hr a 6000 km hr 2 v f 120 km hr Solution: v f v 0 at . t v f v 0 0 a t 30 s v f 2 v 0 2 2 as . s v f 2 v 0 2 0 2 a s 792 m Problem 12-6 A freight train travels at v v 0 1 e b 0 t 0 + , where t is the elapsed time. Determine the distance traveled in time t 1 , and the acceleration at this time. 2
Engineering Mechanics - Dynamics Chapter 12 v 0 60 ft s b 1 s t 1 3 s Solution: v t ( ) v 0 1 e b 0 t 0 + , a t ( ) t v t ( ) d d d t ( ) 0 t t v t ( ) ´ µ d d t 1 + , 123.0 ft a t 1 + , 2.99 ft s 2 Problem 12-7 The position of a particle along a straight line is given by s p = at 3 + bt 2 + ct . Determine its maximum acceleration and maximum velocity during the time interval t 0 d t d t f . Given: a 1 ft s 3 b 9 0 ft s 2 c 15 ft s t 0 0 s t f 10 s Solution: s p at 3 bt 2 . ct . v p t s p d d 3 2 2 . c . a p t v p d d 2 t s p d d 2 6 2 b . Since the acceleration is linear in time then the maximum will occur at the start or at the end. We check both possibilities. a max max 6 0 b . 6 f 2 b . / + , a max 42 ft s 2 The maximum velocity can occur at the beginning, at the end, or where the acceleration is zero. We will check all three locations. t cr b 0 3 a t cr 3 s 3 Given:

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Engineering Mechanics - Dynamics Chapter 12 v max max 3 at 0 2 2 bt 0 . c . 3 f 2 2 f . c . / 3 cr 2 2 cr . c . / + , v max 135 ft s * Problem 12-8 From approximately what floor of a building must a car be dropped from an at-rest position so that it reaches a speed v f when it hits the ground? Each floor is a distance h higher than the one below it. (Note: You may want to remember this when traveling at speed v f ) Given: v f 55 mph h 12 ft g 32.2 ft s 2 Solution: a c g v f 2 0 2 a c s . H v f 2 2 a c H 101.124 ft Number of floors N Height of one floor h 12ft N H h N 8.427 N ceil N ( ) The car must be dropped from floor number N 9 Problem 12–9
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## This note was uploaded on 04/13/2008 for the course ME 2580 taught by Professor Vandenbrink during the Spring '08 term at Western Michigan.

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Chapter 12 - Engineering Mechanics Dynamics Chapter 12...

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