all chapter Engineering mechanics dynamics Hibbeler 14th edition in SI units solution manual pdf PDF

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Authors: Russell Hibbeler
Published: Pearson Education 2016
Edition: 14th SI
Pages: 1273
Type: pdf
Size: 140MB
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12–1.

Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m> s2, where t is in seconds. What is the particle’s velocity when t = 6 s, and what is its position when t = 11 s?

SOLUTION a = 2t - 6 dv = a dt L0

v

dv =

L0

t

(2t - 6) dt

v = t 2 - 6t ds = v dt L0

s

s =

ds =

L0

t

(t2 - 6t) dt

t3 - 3t2 3

When t = 6 s, v = 0

Ans.

When t = 11 s, s = 80.7 m

Ans.

Ans: v = 0 s = 80.7 m

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12–2.

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The acceleration of a particle as it moves along a straight line is given by a = 14t 3 - 12 m>s 2, where t is in seconds. If s = 2 m and v = 5 m>s when t = 0, determine the particle’s velocity and position when t = 5 s. Also, determine the total distance the particle travels during this time period.

SOLUTION L5

v

dv =

0 L

t 3

(4 t - 1) dt

v = t4 - t + 5 2 L

s

ds = s =

L0

t

(t4 - t + 5) dt

1 5 1 t - t2 + 5 t + 2 5 2

When t = 5 s, v = 625 m>s

Ans.

s = 639.5 m

Ans.

Since v Z 0 then d = 6 39.5 - 2 = 6 37.5 m

Ans.

Ans: v = 625 m > s s = 639.5 m d = 637.5 m

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12–3.

The velocity of a particle traveling in a straight line is given by v = (6t - 3t2) m > s, where t is in seconds. If s = 0 when t = 0, determine the particle’s deceleration and position when t = 3 s. How far has the particle traveled during the 3-stime interval, and what is its average speed?

SOLUTION v = 6t - 3t 2 a =

dv = 6 - 6t dt

At t = 3 s a = -12 m >s 2

Ans.

ds = v dt L0

s

ds =

L0

t

(6t - 3t2)dt

s = 3t 2 - t 3 At t = 3 s s = 0

Ans.

Since v = 0 = 6t - 3 t 2, when t = 0 and t = 2 s. when t = 2 s, s = 3(2)2 - (2)3 = 4 m sT = 4 + 4 = 8 m

( vsp )avg =

Ans.

sT 8 = 2.67 m>s = t 3

Ans.

@solutionmanual1 3

Ans: a = -12 m>s2 s = 0 sT = 8 m (vsp)avg = 2.67 m > s

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*12–4.

A particle is moving along a straight line such that its position is defined by s = (10t2 + 20) mm, where t is in seconds. Determine (a) the displacement of the particle during the time interval from t = 1 s to t = 5 s, (b) the average velocity of the particle during this time interval, and (c) the acceleration when t = 1 s .

SOLUTION s = 10t2 + 20 (a) s|1 s = 10(1)2 + 20 = 30 mm s|5 s = 10(5)2 + 20 = 270 mm ¢s = 270 - 30 = 240 mm

Ans.

(b) ¢t = 5 - 1 = 4 s vavg = (c) a =

¢s 240 = 60 mm> s = 4 ¢t

d2 s = 20 mm s2 dt2

Ans.

(for all t)

Ans.

Ans: ∆s = 240 mm v avg = 60 mm > s a = 20 mm> s2

@solutionmanual1 4

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12–5.

A particle moves along a straight line such that its position is defined by s = (t2 - 6t + 5) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s.

SOLUTION s = t2 - 6t + 5 v =

ds = 2t - 6 dt

a =

dv = 2 dt

v = 0 when t = 3 s  t=0 = 5 s  t = 3 = -4 s  t=6 = 5 vavg =

0 ∆s = = 0 ∆t 6

( vsp )avg =

sT ∆t

=

Ans.

9 + 9 = 3 m >s 6

Ans.

a  t = 6 = 2 m>s2

Ans.

Ans: vavg = 0 (v sp) avg = 3 m > s a t = 6 s = 2 m> s2

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12–6.

FOLFNKHUHWRGRZQORDG

A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later.

SOLUTION + T s = s1 + v1 t +

1 ac t 2 2

1 sA = 0 + 0 + ( 9. 81)(2)2 2 sA = 19.62 m 1 sA = 0 + 0 + (9.81)(1)2 2 sB = 4.91 m ¢s = 19.62 - 4.91 = 14.71 m

Ans.

Ans: s = 14.71 m

@solutionmanual1 6

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12–7.

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A bus starts from rest with a constant acceleration of 1 m>s2 . Determine the time required for it to attain a speed of 25 m> s and the distance traveled.

SOLUTION Kinematics: v0 = 0, v = 25 m>s, s0 = 0, and ac = 1 m>s2. + B A:

v = v0 + act 25 = 0 + (1)t t = 25 s

+ B A:

Ans.

v2 = v02 + 2ac(s - s0) 252 = 0 + 2(1)(s - 0) s = 312.5 m

Ans.

Ans: t = 25 s s = 312.5 m

@solutionmanual1 7

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*12–8.

A particle travels along a straight line with a velocity v = (12 - 3t 2) m>s, where t is in seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Determine the acceleration when t = 4 s, the displacement from t = 0 to t = 10 s, and the distance the particle travels during this time period.

SOLUTION v = 12 - 3t 2 a =

dv = -6t dt

s

L-10

(1)

ds =

L 1

t=4

t

v dt =

= -24 m>s2 L1

Ans.

t

( 12 - 3t 2 )dt

s + 10 = 12t - t 3 - 11 s = 12t - t 3 - 21 s s

t =0

t =10

= -21 = -901

∆s = -901 - ( -21) = -880 m

Ans.

From Eq. (1): v = 0 when t = 2s s

t =2

= 12(2) - (2)3 - 21 = -5

sT = (21 - 5) + (901 - 5) = 912 m

Ans.

Ans: a = -24 m > s 2 ∆s = -880 m sT = 912 m

@solutionmanual1 8

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12–9.

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Wh en two cars A and B are next to one another, they are traveling in the same direction with speeds vA and vB , respectively. If B maintains its constant speed, while A begins to decelerate at aA , determine the distance d between the cars at the instant A stops.

A

B

d

SOLUTION Motion of car A: v = v0 + act 0 = vA - aAt

vA

t =

aA

v2 = v02 + 2ac(s - s0) 0 = vA2 + 2( - aA)(sA - 0) sA =

v2A 2aA

Motion of car B: sB = vBt = vB a

vAvB vA b = aA aA

The distance between cars A and B is sBA = |sB - sA| = `

vAvB aA

-

2 2 vA 2vAvB - vA ` = ` ` 2aA 2aA

Ans.

Ans: S BA = `

@solutionmanual1 9

2 vA vB - v2A ` 2aA

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12–10.

A particle travels along a straight-line path such that in 4 s it moves from an initial position sA = -8 m to a position sB = +3 m. Then in another 5 s it moves from sB to sC = -6 m. Determine the particle’s average velocity and average speed during the 9-s time interval.

SOLUTION Average Velocity: The displacement from A to C is ∆s = sC - SA = -6 - ( -8) = 2 m. ∆s 2 vavg = = = 0.222 m>s Ans. ∆t 4 + 5 Average Speed: The distances traveled from A to B and B to C are sASB = 8 + 3 = 11.0 m and sBSC = 3 + 6 = 9.00 m, respectively. Then, the total distance traveled is sTot = sASB + sBSC = 11.0 + 9.00 = 20.0 m. (vsp)avg =

sTot 20.0 = = 2.22 m>s 4 + 5 ∆t

Ans.

Ans:

@solutionmanual1 10

vavg = 0.222 m > s (vsp)avg = 2.22 m > s

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12–11.

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Traveling with an initial speed of 70 km >h, a car accelerates at 6000 km>h2 along a straight road. How long will it take to reach a speed of 120 km> h? Also, through what distance does the car travel during this time?

SOLUTION v = v1 + ac t 120 = 70 + 6000(t) t = 8.33(10 - 3) hr = 30 s

Ans.

v2 = v21 + 2 ac(s - s1) (120)2 = 702 + 2(6000)(s - 0) s = 0.792 km = 792 m

Ans.

Ans: t = 30 s s = 792 m

@solutionmanual1 11

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*12–12.

A particle moves along a straight line with an acceleration of a = 5> (3s1>3 + s5>2) m>s2, where s is in meters. Determine the particle’s velocity when s = 2 m , if it starts from rest when s = 1 m .Use a numerical method to evaluate the integral.

SOLUTION a =

5 5

1

A 3s3 + s 2 B

a ds = v dv 2

L1 A 3s + s B 5 ds 1 3

0.8351 =

5 2

=

L0

v

v dv

1 2 v 2

v = 1.29 m> s

Ans .

Ans: v = 1.29 m> s

@solutionmanual1 12

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12–13.

The acceleration of a particle as it moves along a straight line is given by a = (2t - 1) m>s 2, where t is in seconds. If s = 1 m and v = 2 m > s when t = 0, determine the particle’s velocity and position when t = 6 s. Also, determine the total distance the particle travels during this time period.

SOLUTION a = 2t - 1 dv = a dt L2

v

L0

dv =

t

(2t - 1)dt

v = t2 - t + 2 dx = v dt Lt

s

ds =

s =

L0

t

(t2 - t + 2)dt

1 3 1 t - t 2 + 2t + 1 3 2

When t = 6 s v = 32 m > s

Ans.

s = 67 m

Ans.

Since v ≠ 0 for 0 … t … 6 s, then d = 67 - 1 = 66 m

Ans.

Ans: v = 32 m > s s = 67 m d = 66 m

@solutionmanual1 13

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12–14.

A train starts from rest at station A and accelerates at 0.5 m>s2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 1 m>s2 until it is brought to rest at station B. Determine the distance between the stations.

SOLUTION Kinematics: For stage (1) motion, v0 = 0, s0 = 0, t = 60 s, and ac = 0.5 m>s2. Thus, + B A:

s = s0 + v0t + s1 = 0 + 0 +

+ B A:

1 a t2 2 c

1 (0.5)(602) = 900 m 2

v = v0 + act v1 = 0 + 0.5(60) = 30 m>s

For stage (2) motion, v0 = 30 m>s, s0 = 900 m, ac = 0 and t = 15(60) = 900 s.Thus, + B A:

s = s0 + v0t +

1 act2 2

s2 = 900 + 30(900) + 0 = 27 900 m For stage (3) motion, v0 = 30 m>s, v = 0, s0 = 27 900 m and ac = -1 m>s2. Thus, + A: B

v = v0 + act 0 = 30 + (-1)t t = 30 s

+ :

s = s0 + v0t +

1 2 act 2

s3 = 27 900 + 30(30) +

1 (-1)(302) 2 Ans.

= 28 350 m = 28.4 km

Ans: s = 28.4 km

@solutionmanual1 14

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12–15.

FOLFNKHUHWRGRZQORDG

A particle is moving along a straight line such that its velocity is defined as v = (-4s2) m> s, where s is in meters. If s = 2 m when t = 0, determine the velocity and acceleration as functions of time.

SOLUTION v = -4s2 ds = -4s2 dt 2 L

s

s - 2 ds =

L0

t

-4 dt

-s - 1| s2 = -4t|t0 t =

1 -1 (s - 0.5) 4

s =

2 8t + 1

v = -4a

a =

2 16 2 b = m> s 8t + 1 (8t + 1)2

Ans.

16(2)(8t + 1)(8) 256 dv = = m>s2 (8t + 1)4 (8t + 1)3 dt

Ans.

Ans: 16 m >s (8t + 1)2 256 a = m >s2 (8t + 1)3 v =

@solutionmanual1 15

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*12–16.

Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate at 1.5 m>s2 and decelerate at 2 m>s2.

SOLUTION Using formulas of constant acceleration: v2 = 1.5 t1 x =

1 (1.5)(t12) 2

0 = v2 - 2 t2 1 (2)(t22) 2

1000 - x = v2t2 -

Combining equations: t1 = 1.33 t2; v2 = 2 t2 x = 1.33 t22 1000 - 1.33 t22 = 2 t22 - t22 t2 = 20.702 s;

t1 = 27.603 s

t = t1 + t2 = 48.3 s

Ans.

Ans: t = 48.3 s

@solutionmanual1 16

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FOLFNKHUHWRGRZQORDG

12–17.

A parti cle is moving with a veloci ty of v0 wh en s = 0 and t = 0. If it is subjected to a deceleration of a = -kv3, where k is a constant, determine its velocity and position as functions of time.

SOLUTION a =

dn = - kn3 dt

n

Ln0

n - 3 dn =

L 0

t

- k dt

1 -2 1n - n0- 22 = - kt 2

-

n = a 2kt + a

1

-2 1 2b b n0

Ans.

ds = n dt L0

s

ds =

L0

t

dt a 2kt + a

2a 2kt + a s =

2k

1

2 1 2b b v0

1

t 2 1 2b b 3 n0

0

1 1 1 B ¢2kt + ¢ 2 ≤ ≤ - R n0 k n0 1 2

s =

Ans.

Ans: - 1>2

b v20 1 1 1 1>2 d s = c a2kt + 2 b v0 k v0 v = a2kt +

@solutionmanual1 17

1

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FOLFNKHUHWRGRZQORDG

12–18.

A particle is moving a long a straigh t line w ith an initial velocity of 6 m>s when it is subjected to a deceleration of a = (-1.5v1>2) m>s2 , where v is in m> s . Determine how far it travels before it stops. How much time does this take?

SOLUTION Distance Traveled: The distance traveled by the particle can be determined by applying Eq. 12–3. vdv a

ds = L0

s

v

ds =

v

L6 m>s -1.5v 2 1

dv

v

s =

L6 m>s

1

-0.6667 v2 dv

= a -0.4444v 2 + 6.532 b m 3

When v = 0,

s = -0.4444 a 0 2 b + 6.532 = 6.53 m 3

Ans.

Time: The time required for the particle to stop can be determined by applying Eq. 12–2. dt = L0

t

v

dt = -

t = -1.333 a v 2 b 1

When v = 0,

dv a

v 6 m>s

6 m>s L

dv 1

1.5v 2

= a 3.266 - 1.333v 2 b s 1

t = 3.266 - 1.333 a 0 2 b = 3.27 s 1

Ans.

Ans: s = 6.53 m t = 3.27 s

@solutionmanual1 18

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