Solution Manual For Engineering Mechanics Bedford

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Hi im looking for the solutions manual for engineering mechanics dynamics 5th edition by bedford fowler can you send it to me at johntpoweralstomcom find all.

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Problem 13.1 In Example 13.2, suppose that the vehicle is dropped from a height h = 6m. (a) What is the downward velocity 1 s after it is released? (b) What is its downward velocity just before it reaches the ground? Solution: The equations that govern the motion are: a = −g = −9.81 m/s2 v = −gt h s = − 21 gt 2 + h (a) (b) v = −gt = −(9.81 m/s2 )(1 s) = −9.81 m/s. The downward velocity is 9.81 m/s. We need to first determine the time at which the vehicle hits the ground   2h 2(6 m) 1 2 = 1.106 s = s = 0 = − 2 gt + h ⇒ t = g 9.81 m/s2 Now we can solve for the velocity v = −gt = −(9.81 m/s2 )(1.106 s) = −10.8 m/s. The downward velocity is 10.8 m/s.

Problem 13.2 The milling machine is programmed so that during the interval of time from t = 0 to t = 2 s, the position of its head (in inches) is given as a function of time by s = 4t − 2t 3. What are the velocity (in in/s) and acceleration (in in/s2 ) of the head at t = 1 s?

Solution: The motion is governed by the equations s = (4 in/s)t − (2 in/s2 )t 2, s v = (4 in/s) − 2(2 in/s2 )t, a = −2(2 in/s2 ). At t = 1 s, we have v = 0, a = −4 in/s2. C 2008 Pearson Education South Asia Pte Ltd.

All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior  to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise. Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. 9 Problem 13.3 In an experiment to estimate the acceleration due to gravity, a student drops a ball at a distance of 1 m above the floor. His lab partner measures the time it takes to fall and obtains an estimate of 0.46 s. S (a) What do they estimate the acceleration due to gravity to be?

(b) Let s be the ball’s position relative to the floor. Using the value of the acceleration due to gravity that they obtained, and assuming that the ball is released at t = 0, determine s (in m) as a function of time. Solution: The governing equations are a = −g v = −gt s0 s = − 12 gt 2 + h (a) When the ball hits the floor we have 0 = − 12 gt 2 + h ⇒ g = 2h 2(1 m) = = 9.45 m/s2 t2 (0.46 s)2 g = 9.45 m/s2 (b) The distance s is then given by s = − 12 (9.45 m/s2 ) + 1 m.

S = −(4.73 m/s2 )t 2 + 1.0 m. Problem 13.4 The boat’s position during the interval of time from t = 2 s to t = 10 s is given by s = 4t + 1.6t 2 − 0.08t 3 m. (a) Determine the boat’s velocity and acceleration at t = 4 s.

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(b) What is the boat’s maximum velocity during this interval of time, and when does it occur? Solution: s = 4t + 1.6t 2 − 0.08t 3 10 a) v(4s) = 12.96 m/s2 v= a(4s) = 1.28 m/s2 ds = 4 + 3.2t − 0.24t 2 ⇒ dt b) a = 3.2 − 0.48t = 0 ⇒ t = 6.67s a= dv = 3.2 − 0.48t dt v(6.67s) = 14.67 m/s c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior  to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise. Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. Problem 13.7 The position of a point during the interval of time from t = 0 to t = 3 seconds is s = 12 + 5t 2 − t 3 m.

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(a) (b) What is the maximum velocity during this interval of time, and at what time does it occur? What is the acceleration when the velocity is a maximum? Solution: (a) This is indeed a maximum, since ds = 10t − 3t 2.

The maximum occurs when The velocity is dt dv = 10 − 6t = 0, from which dt t= 10 = 1.667 seconds. 6 velocity is d2v = −6 1 s a = −c (constant), v = −ct + 13.33 m/s, s = −c t2 + ( 13.33 m/s )t + 13.33 m 2 At the stop we have 90 m = −c t2 + (13.33 m/s)t + 13.33 m 2 0 = −ct + 1 3. 3 3 m /s ⇒ a) c = 1.1 7 m /s 2 b) t = 11.41 s Problem 13.30 The car is traveling at 48 km/h when the traffic light 90 m ahead turns yellow. The driver takes 1 s to react before he applies the accelerator. If the car has a constant acceleration of 2 m/s2 and the light remains yellow for 5 s, will the car reach the light before it turns red? How fast is the car moving when it reaches the light?

Solution: First, convert the initial speed into m/s. 48 km/h = 13.33 m/s. At the end of the 5 s, the car will have traveled a distance   1 (2 m/s 2 )(5 s − 1 s)2 + ( 13.33 m/s )(5 s − 1 s) = 82.65 m.

D = (13.33 m/s )(1 s) + 2 When the light turns red, the driver will still be 7.35 m from the light. To find the time at which the car does reach the light, we solve   1 (2 m/s2 )(t − 1 s)2 + (13.33 m/s )(t − 1 s) 90 m = (13.33 m/s )(1 s) + 2 ⇒ t = 5.34 s. The speed at this time is v = 13.33 m/s + ( 2 m/s 2 ) (5.34 s − 1 s) = 22.01 m/s.

V = 79.2 k m/ h. C 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior  to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.

Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. 21 Problem 13.31 A high-speed rail transportation system has a top speed of 100 m/s. For the comfort of the passengers, the magnitude of the acceleration and deceleration is limited to 2 m/s2. Determine the time required for a trip of 100 km.

Engineering Mechanics Dynamics Solution Manual

Strategy: A graphical approach can help you solve this problem. Recall that the change in the position from an initial time t0 to a time t is equal to the area defined by the graph of the velocity as a function of time from t0 to t. Solution: Divide the time of travel into three intervals: The time required to reach a top speed of 100 m/s, the time traveling at top speed, and the time required to decelerate from top speed to zero. From symmetry, the first and last time intervals are equal, and the distances traveled during these intervals are equal.