# Dynamics Solution

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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

•12–1. A car starts from rest and with constant acceleration achieves a velocity of 15 m>s when it travels a distance of 200 m. Determine the acceleration of the car and the time required.

Kinematics: v0 = 0, v = 15 m>s, s0 = 0, and s = 200 m. + A:B v2 = v0 2 + 2ac(s - s0) 152 = 02 + 2ac(200 - 0) ac = 0.5625 m>s2 + A:B v = v0 + act 15 = 0 + 0.5625t t = 26.7 s Ans. Ans.

12–2. A train starts from rest at a station and travels with a constant acceleration of 1 m>s2. Determine the velocity of the train when t = 30 s and the distance traveled during this time.

Kinematics: ac = 1 m>s2, v0 = 0, s0 = 0, and t = 30 s. + A:B + A:B v = v0 + act = 0 + 1(30) = 30 m>s s = s0 + v0t + 1 2 at 2 c Ans.

= 0 + 0 + = 450 m

1 (1) A 302 B 2 Ans.

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12–3. An elevator descends from rest with an acceleration of 5 ft>s2 until it achieves a velocity of 15 ft>s. Determine the time required and the distance traveled.

Kinematics: ac = 5 ft>s2, v0 = 0, v = 15 ft>s, and s0 = 0.

A+TB

v = v0 + act 15 = 0 + 5t t = 3s Ans.

A+TB

v2 = v0 2 + 2ac(s - s0) 152 = 02 + 2(5)(s - 0) s = 22.5 ft Ans.

*12–4. A car is traveling at 15 m>s, when the traffic light 50 m ahead turns yellow. Determine the required constant deceleration of the car and the time needed to stop the car at the light.

Kinematics: v0 = 0, s0 = 0, s = 50 m and v0 = 15 m>s. + A:B v2 = v0 2 + 2ac(s - s0) 0 = 152 + 2ac(50 - 0) ac = -2.25 m>s2 = 2.25 m>s2 ; + A:B v = v0 + act 0 = 15 + (-2.25)t t = 6.67 s Ans. Ans.

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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

•12–5. A particle is moving along a straight line with the acceleration a = (12t – 3t1/2) ft>s2, where t is in seconds. Determine the velocity and the position of the particle as a function of time. When t = 0, v = 0 and s = 15 ft.

Velocity: + A:B dv = a dt

v 0 L v t 0 L

dv =

v 0 = A 6t2 - 2t3>2 B 2 + A:B

A 12t - 3t1>2 B dt

t 0

v = A 6t2 - 2t3>2 B ft>s

Ans.

Position: Using this result and the initial condition s = 15 ft at t = 0 s, ds = v dt s 15 L ft s t

ds =

0 L

A 6t2 - 2t3>2 B dt

4 5>2 2 t t b 5 0

s 15 ft = a 2t3 s = a 2t3 -

4 5>2 t + 15b ft 5

Ans.

12–6. A ball is released from the bottom of an elevator which is traveling upward with a velocity of 6 ft>s. If the ball strikes the bottom of the elevator shaft in 3 s, determine the height of the elevator from the bottom of the shaft at the instant the ball is released. Also, find the velocity of the ball when it strikes the bottom of the shaft. Kinematics: When the ball is released, its velocity will be the same as the elevator at the instant of release. Thus, v0 = 6 ft>s. Also, t = 3 s, s0 = 0, s = -h, and ac = -32.2 ft>s2.

A+cB

s = s0 + v0t +

1 a t2 2 c 1 (-32.2) A 32 B 2 Ans.

-h = 0 + 6(3) + h = 127 ft

A+cB

v = v0 + act v = 6 + (-32.2)(3) = -90.6 ft>s = 90.6 ft>s T Ans.

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