Hibbeler 14th Dynamics solution Manual PDF

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INSTRUCTOR SOLUTIONS MANUAL

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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

ds =

s =

L0

t

(t2 - 6t) dt

t3 - 3t2 3

When t = 6 s, Ans.

v = 0 When t = 11 s,

Ans.

s = 80.7 m

Ans: s = 80.7 m 1

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12–2. If a particle has an initial velocity of v0 = 12 ft>s to the right, at s0 = 0, determine its position when t = 10 s, if a = 2 ft>s2 to the left.

SOLUTION +2 s = s0 + 1S

v0 t +

1 2 a t 2 c

= 0 + 12(10) +

1 ( - 2)(10)2 2 Ans.

= 20 ft

Ans: s = 20 ft 2

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12–3. 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

(1)

dv = - 6t dt

a = s

t=4

= -24 m>s2

t

ds =

L-10

L1

v dt =

t

L1

Ans.

( 12 - 3t 2 ) dt

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

t=0

s

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 Ans.

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

Ans: a = - 24 m>s2 ∆s = - 880 m sT = 912 m 3

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*12–4. A particle travels along a straight line with a constant acceleration. When s = 4 ft, v = 3 ft>s and when s = 10 ft, v = 8 ft>s. Determine the velocity as a function of position.

SOLUTION Velocity: To determine the constant acceleration ac, set s0 = 4 ft, v0 = 3 ft>s, s = 10 ft and v = 8 ft>s and apply Eq. 12–6. + ) (:

v2 = v20 + 2ac (s - s0) 82 = 32 + 2ac (10 - 4) ac = 4.583 ft>s2

Using the result ac = 4.583 ft>s2, the velocity function can be obtained by applying Eq. 12–6. v2 = v20 + 2ac (s - s0) v2 = 32 + 2(4.583) (s - 4)

A

+ ) (:

v = A 29.17s - 27.7 ft>s

Ans.

Ans: v = 4

( 19.17s - 27.7 ) ft>s

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12–5. 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>s2

Ans.

ds = v dt L0

s

ds =

L0

t

(6t - 3t2)dt

s = 3t 2 - t 3 At t = 3 s Ans.

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

Ans.

sT = 4 + 4 = 8 m

( vsp ) avg =

sT 8 = = 2.67 m>s t 3

Ans.

Ans: sT = 8 m vavg = 2.67 m>s 5

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12–6. The position of a particle along a straight line is given by s = (1.5t3 - 13.5t2 + 22.5t) ft, where t is in seconds. Determine the position of the particle when t = 6 s and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled.

SOLUTION Position: The position of the particle when t = 6 s is s|t = 6s = 1.5(63) - 13.5(62) + 22.5(6) = - 27.0 ft

Ans.

Total DistanceTraveled: The velocity of the particle can be determined by applying Eq. 12–1. v =

ds = 4.50t2 - 27.0t + 22.5 dt

The times when the particle stops are 4.50t2 - 27.0t + 22.5 = 0 t = 1s

and

t = 5s

The position of the particle at t = 0 s, 1 s and 5 s are s t = 0 s = 1.5(03) - 13.5(02) + 22.5(0) = 0 s t = 1 s = 1.5(13) - 13.5(12) + 22.5(1) = 10.5 ft s t = 5 s = 1.5(53) - 13.5(52) + 22.5(5) = - 37.5 ft From the particle’s path, the total distance is Ans.

stot = 10.5 + 48.0 + 10.5 = 69.0 ft

Ans: s t = 6 s = - 27.0 ft stot = 69.0 ft 6

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12–7. 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 =

∆s 0 = = 0 ∆t 6

( vsp ) avg =

Ans.

sT 9 + 9 = = 3 m>s ∆t 6

Ans.

a  t = 6 = 2 m>s2

Ans.

Ans: vavg = 0 (vsp)avg = 3 m>s a  t = 6 s = 2 m>s2 7

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*12–8. 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 Ans.

¢s = 270 - 30 = 240 mm (b) ¢t = 5 - 1 = 4 s vavg = (c) a =

240 ¢s = = 60 mm>s ¢t 4

d2s = 20 mm s2 dt2

Ans. Ans.

(for all t)

Ans: ∆s = 240 mm vavg = 60 mm>s a = 20 mm>s2 8

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12–9. The acceleration of a particle as it moves along a straight line is given by a = (2t - 1) m>s2, 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

t

L0

dv =

(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 Ans.

d = 67 - 1 = 66 m

Ans: v = 32 m>s s = 67 m d = 66 m 9

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12–10. A particle moves along a straight line with an acceleration of a = 5>(3s1>3 + s 5>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 1 3

5

A 3s + s2 B

a ds = v dv 2

v

5 ds 1 3

L1 A 3s + s 0.8351 =

5 2

B

=

L0

v dv

1 2 v 2 Ans.

v = 1.29 m>s

Ans: v = 1.29 m>s 10

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12–11. 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 sA S B = 8 + 3 = 11.0 m and sB S C = 3 + 6 = 9.00 m, respectively. Then, the total distance traveled is sTot = sA S B + sB S C = 11.0 + 9.00 = 20.0 m. (vsp)avg =

sTot 20.0 = = 2.22 m>s ∆t 4 + 5

Ans.

Ans: vavg = 0.222 m>s (vsp)avg = 2.22 m>s 11

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*12–12. 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) Ans.

s = 0.792 km = 792 m

Ans: t = 30 s s = 792 m 12

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12–13. Tests reveal that a normal driver takes about 0.75 s before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0.1% alcohol in his system to do the same. If such drivers are traveling on a straight road at 30 mph (44 ft>s) and their cars can decelerate at 2 ft>s2, determine the shortest stopping distance d for each from the moment they see the pedestrians. Moral: If you must drink, please don’t drive!

v1

44 ft/s

d

SOLUTION Stopping Distance: For normal driver, the car moves a distance of d¿ = vt = 44(0.75) = 33.0 ft before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with s0 = d¿ = 33.0 ft and v = 0. + B A:

v2 = v20 + 2ac (s - s0) 02 = 442 + 2(- 2)(d - 33.0) Ans.

d = 517 ft

For a drunk driver, the car moves a distance of d¿ = vt = 44(3) = 132 ft before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with s0 = d¿ = 132 ft and v = 0. + B A:

v2 = v20 + 2ac (s - s0) 02 = 442 + 2(- 2)(d - 132) Ans.

d = 616 ft

Ans: Normal: d = 517 ft drunk: d = 616 ft 13

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12–14. The position of a particle along a straight-line path is defined by s = (t3 - 6t2 - 15t + 7) ft, where t is in seconds. Determine the total distance traveled when t = 10 s. What are the particle’s average velocity, average speed, and the instantaneous velocity and acceleration at this time?

Solution s = t3 - 6t 2 - 15t + 7 v =

ds = 3t2 - 12t - 15 dt

When t = 10 s, Ans.

v = 165 ft>s a =

dv = 6t - 12 dt

When t = 10 s, a = 48 ft>s2

Ans.

When v = 0, 0 = 3t 2 - 12t - 15 The positive root is t = 5s When t = 0,  s = 7 ft When t = 5 s,  s = - 93 ft When t = 10 s,  s = 257 ft Total distance traveled sT = 7 + 93 + 93 + 257 = 450 ft

Ans.

∆s 257 - 7 = = 25.0 ft>s ∆t 10 - 0

Ans.

sT 450 = = 45.0 ft>s ∆t 10

Ans.

vavg =

( vsp ) avg =

Ans: v = 165 ft>s a = 48 ft>s2 sT = 450 ft vavg = 25.0 ft>s (vsp)avg = 45.0 ft>s 14

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12–15. A particle is moving with a velocity of v0 when 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 dn = - kn3 dt

a =

t

n

n - 3 dn =

Ln0

L0

- k dt

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

-

n = a 2kt + a

1

-2 1 b b n20

Ans.

ds = n dt s

L0

s = s =

t

ds =

L0

dt a 2kt + a

2 a 2kt + a 2k

1

2 1 bb v20

1

t 2 1 b b 3 n20

0

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

Ans.

Ans: v = a2kt + s=

15

1 - 1>2 b v20

1 1 1>2 1 c a2kt + 2 b d k v0 v0

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*12–16. A particle is moving along a straight line with 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. ds =

vdv a

s

L0

v

ds =

v 1

L6 m>s - 1.5v2

v

s =

dv

1

L6 m>s

- 0.6667 v2 dv 3

= a -0.4444v2 + 6.532 b m When v = 0,

3

Ans.

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

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

dv a

t

L0

v

dt = 1

t = - 1.333a v2 b When v = 0,

v 6 m>s

dv 1

L6 m>s 1.5v 2 1

= a3.266 - 1.333v 2 b s 1

Ans.

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

Ans: s = 6.53 m t = 3.27 s 16

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12–17. Car B is traveling a distance d ahead of car A. Both cars are traveling at 60 ft>s when the driver of B suddenly applies the brakes, causing his car to decelerate at 12 ft>s2 . It takes the driver of car A 0.75 s to react (this is the normal reaction time for drivers). When he applies his brakes, he decelerates at 15 ft>s2. Determine the minimum distance d be tween the cars so as to avoid a collision.

A

B d

SOLUTION For B: + ) (:

v = v0 + ac t vB = 60 - 12 t

+ ) (:

s = s0 + v0 t +

1 a t2 2 c

sB = d + 60t -

1 (12) t2 2

(1)

For A: + ) (:

v = v0 + ac t

vA = 60 - 15(t - 0.75), [t 7 0.75] + ) (:

s = s0 + v0 t +

1 a t2 2 c

sA = 60(0.75) + 60(t - 0.75) -

1 (15) (t - 0.75)2, 2

[t 7 0.74]

(2)

Require vA = vB the moment of closest approach. 60 - 12t = 60 - 15(t - 0.75) t = 3.75 s Worst case without collision would occur when sA = sB. At t = 3.75 s, from Eqs. (1) and (2): 60(0.75) + 60(3.75 - 0.75) - 7.5(3.75 - 0.75)2 = d + 60(3.75) - 6(3.75)2 157.5 = d + 140.625 Ans.

d = 16.9 ft

Ans: d = 16.9 ft 17

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12–18. The acceleration of a rocket traveling upward is given by a = 16 + 0.02s2 m>s2, where s is in meters. Determine the time needed for the rocket to reach an altitude of s = 100 m. Initially, v = 0 and s = 0 when t = 0.

SOLUTION a ds = n dv

s

s

L0

16 + 0.02 s2 ds =

6 s + 0.01 s2 =

n

L0

n dn

1 2...