Chapter 12

Download Document
Showing pages : 1 - 5 of 144
This preview has blurred sections. Sign up to view the full version! View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 t a 1.852 m s 2 d v 1 t 1 2 a t 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 a d a v 2 2 d a 6.4 ft s 2 v a t 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 . 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 18 ft s Solution: v v 2 2 g h . v 59.5 ft s 1 Engineering Mechanics - Dynamics Chapter 12 t v v g t 1.29 s * Problem 124 Starting from rest, a particle moving in a straight line has an acceleration of a = ( bt + c ). What is the particles velocity at t 1 and what is its position at t 2 ? Given: b 2 m s 3 c 6 m s 2 t 1 6 s t 2 11 s Solution: a t ( ) b t c . v t ( ) t t a t ( ) ´ µ ¶ d d t ( ) t t v t ( ) ´ µ ¶ d v t 1 + , m s d t 2 + , 80.7 m Problem 12-5 Traveling with an initial speed v 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 70 km hr a 6000 km hr 2 v f 120 km hr Solution: v f v a t . t v f v a t 30 s v f 2 v 2 2 a s . s v f 2 v 2 2 a s 792 m Problem 12-6 A freight train travels at v v 1 e b t + , 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 60 ft s b 1 s t 1 3 s Solution: v t ( ) v 1 e b t + , a t ( ) t v t ( ) d d d t ( ) 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 d t d t f . Given: a 1 ft s 3 b 9 ft s 2 c 15 ft s t 0 s t f 10 s Solution: s p a t 3 b t 2 . c t . v p t s p d d 3 a t 2 2 b t . c . a p t v p d d 2 t s p d d 2 6 a t 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 a t b . 6 a t 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 3 a t cr 3 s 3 Given: Engineering Mechanics - Dynamics Chapter 12 v max max 3 a t 2 2 b t . c . 3 a t f 2 2 b t f . c . / 3 a t cr 2 2 b t 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 one below it....
View Full Document