BeerVectorDynamicsISM_Ch11_v2

PROBLEM 11.1

A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS coordinates are used to determine his displacement as a function of time: x= 0.5t3 + t2 + 2t where x and t are expressed in ft and seconds, respectively. Determine the position, velocity, and acceleration of the boarder when t = 5 seconds. SOLUTION Position: x0.5t3 t2 2t Velocity: v dx 1.5t2 2t 2 dt     Acceleration: a dv 3t 2 dt    At t5 s, x0.5 5

 

3 52 2 5

 

x97.5 ft

 

2

 

1.5 5 2 5 2 v   v49.5 ft/s

 

3 5 2 a   a17 ft/s2

The motion of a particle is defined by the relation _23_92_{12 10,}

x t t t where x and t are expressed in feet

and seconds, respectively. Determine the time, the position, and the acceleration of the particle when v  0.

SOLUTION 3 2 2 9 12 10 x t  t  t Differentiating, 62 18 12 6( 2 3 2) 6( 2)( 1)           dx v t t t t dt t t 12 18 dv  a t dt So v0 at t1s and t2 s. At t1 s, x₁  2 9 12 10 15  t1.000 s  112 18  6 a x₁15.00 ft  2 1 6.00 ft/s a  At t2 s, 3 2 2 2(2) 9(2) 12(2) 10 14  x t2.00 s  214.00 ft x  2 (12)(2) 18 6 a    a₂6.00 ft/s2 

PROBLEM 11.3

The vertical motion of mass A is defined by the relation x10 sin 2t15cos 2t100, where x and t are expressed in mm and seconds, respectively. Determine (a) the position, velocity and acceleration of A when t  1 s, (b) the maximum velocity and acceleration of A.

SOLUTION 10sin 2 15cos 2 100    x t t 20cos 2 30sin 2 dx  v t t dt 40sin 2 60cos 2 dv   a t t dt

For trigonometric functions set calculator to radians:

(a) At 1 s.t x₁10sin 2 15cos 2 100 102.9   x₁102.9 mm  1 20cos 2 30sin 2 35.6

v     v₁ 35.6 mm/s 

1 40sin 2 60cos 2  11.40

a a₁ 11.40 mm/s2 

(b) Maximum velocity occurs when a 0.

40sin 2 t60cos 2t 0 tan 2 60 1.5

40

t   

2ttan ( 1.5)1   0.9828 and 0.9828   Reject the negative value. 2t2.1588

t1.0794 s

t1.0794 s for vmax

max 20cos(2.1588) 30sin(2.1588) 36.056

v  

  v_max  36.1 mm/s 

Note that we could have also used

2 2

max 20 30 36.056

v   

by combining the sine and cosine terms.

For amax we can take the derivative and set equal to zero or just combine the sine and cosine terms.

2 2 2

max 40 60 72.1 mm/s

a    a_max 72.1 mm/s2 

A loaded railroad car is rolling at a constant velocity when it couples with a spring and dashpot bumper system. After the coupling, the motion of the car is defined by the relation 60 4.8tsin16

x e t where x and t

are expressed in mm and seconds, respectively. Determine the position, the velocity and the acceleration of the railroad car when (a) t  0, (b) t  0.3 s. SOLUTION 4.8 60 tsin16 x e t 4.8 4.8 4.8 4.8 60( 4.8) sin16 60(16) cos16 288 sin16 960 cos16 t t t t dx v e t e t dt v e t e t            4.8 4.8 4.8 4.8 4.8 4.8 1382.4 sin16 4608 cos16 4608 cos16 15360 sin16 13977.6 sin16 9216 cos16 t t t t t dv a e t e t dt e t e t a e t e t               (a) At 0,t x₀ 0 x₀0 mm  0 960 mm/s v  v₀ 960 mm/s  a₀  9216 mm/s2 a₀9220 mm/s2  (b) At 0.3 s,t 4.8 1.44 0.23692 sin16 sin 4.8 0.99616 cos16 cos 4.8 0.08750 t e e t t  _  _      0.3 (60)(0.23692)( 0.99616) 14.16 x     x_0.314.16 mm  0.3 (288)(0.23692)( 0.99616) (960)(0.23692)(0.08750) 87.9 v      v_0.3 87.9 mm/s  0.3 (13977.6)(0.23692)( 0.99616) (9216)(0.23692)(0.08750) 3108 a      a_0.33110 mm/s2  or 3.11 m/s 2 

PROBLEM 11.5

The motion of a particle is defined by the relation x6t42t312t2   where x and t are expressed in 3t 3, meters and seconds, respectively. Determine the time, the position, and the velocity when a 0.

SOLUTION We have x6t4 2t312t2   3t 3 Then _v dx _24t3 _6t2 _{24t 3} dt      and a dv 72t2 12t 24 dt     When a 0: 72t212t24 12(6 t2 t 2) 0 or (3t2)(2t  1) 0 or 2s and 1s (Reject) 3 2 t t  t0.667 s  At 2 s: 3 t 4 3 2 2/3 2 2 2 2 6 2 12 3 3 3 3 3 3 x   _{ }   _{ }   _{ }   _{ }         or x2/30.259 m  3 2 2/3 2 2 2 24 6 24 3 3 3 3 v   _{ }   _{ }   _{ }       or v2/3 8.56 m/s 

The motion of a particle is defined by the relation x t3 9t224t where x and t are expressed in inches 8, and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.

SOLUTION We have x t3 9t2 24t 8 Then _v dx _3t2 _{18t 24} dt     and a dv 6t 18 dt    (a) When v 0: 3t218t24 3( t2   6t 8) 0 (t2)(t4) 0 t2.00 s and t4.00 s  (b) When a 0: 6t18 0 or t3 s At t3 s: x₃ (3)39(3)224(3) 8 or x₃10.00 in. 

First observe that 0 t 2 s: v 0

2 s t 3 s: v 0 Now At 0:t x₀  8 in. At 2 t s: x₂ (2)39(2)224(2) 8 12 in.  Then _{2 0} 3 2 12 ( 8) 20 in. | | |10 12| 2 in. x x x x         

PROBLEM 11.7

A girl operates a radio-controlled model car in a vacant parking lot. The girl’s position is at the origin of the xy coordinate axes, and the surface of the parking lot lies in the x-y plane. She drives the car in a straight line so that the x coordinate is defined by the relation x(t) = 0.5t3 - 3t2 + 3t + 2, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and total distance travelled when the acceleration is zero. SOLUTION Position:

x t

 



0.5

t

3



3

t

2

 

3

t

2

Velocity: v t

 

dx dt 

 

_1.52 _{6 3} v t  t  t

(a) Time when v=0 0 1.5 t2  6t 3

  

2 6 6 4 1.5 3 2 1.5 t    t0.586 s and t3.414 s  Acceleration: a t

 

dv dt 

 

3 6 a t   t

Time when a=0 0 3  t 6 2 s

t

(b) Position at t=2 s

x

 

2



0.5 2

 

3



3 2

 

2

  

3 2 2

 

2 0 m

x  

To find total distance note that car changes direction at t=0.586 s Position at t=0 s

x

 

0



0.5 0

 

3



3 0

 

2

  

3 0 2

 

0

2

x



Position at t=0.586 s

x



0.586





0.5 0.586





3



3 0.586





2

 

3 0.586 2





0.586



2.828 m

x



Distances traveled: From t=0 to t=0.586 s:

x



0.586

  



x

0



0.828 m

From t=0.586 to t=2 s:

x

  

2



x

0.586





2.828 m

The motion of a particle is defined by the relation x  t2



t 2 ,



3 where x and t are expressed in feet and seconds, respectively. Determine (a) the two positions at which the velocity is zero, (b) the total distance traveled by the particle from t  0 to t  4 s..

SOLUTION Position: x t

 

  t2



t 2



3 Velocity: v t

 

dx dt 

 





2 2 3 2 v t  t t

 



2



2 2 3 4 4 3 14 12 v t t t t t t         Time when v(t)=0 0 3t2 14t12

 



 

2 14 14 4 3 12 2 3 t      1.131 s and 3.535 s t t (a) Position at t=1.131 s

x



1.131

 



1.131

 

2



1.1.31



2



3



1.131 1.935 ft.



x



 Position at t=3.535 s

x



3.535

 



3.535

 

2



3.535



2



3



3.531



8.879 ft.

x





To find total distance traveled note that the particle changes direction at t=1.131 s and again at t=3.535 s. Position at t=0 s

x

    

0



0

2



0



2



3

 

0 8 ft x  Position at t=4 s

x

    

4



4

2



4 2





3

 

4 8 ft x  (b) Distances traveled: From t=0 to t=1.131 s:

x



1.131

  



x

0



6.065 ft.

From t=1.131 to t=3.531 s:

x



3.535

 



x

1.131





6.944 ft.

From t=3.531 to t=4 s:

x

  

4



x

3.535





0.879 ft.

PROBLEM 11.9

The brakes of a car are applied, causing it to slow down at a rate of 10 ft/s2. Knowing that the car stops in 100 ft, determine (a) how fast the car was traveling immediately before the brakes were applied, (b) the time required for the car to stop.

SOLUTION 2 10 ft/s   a (a) Velocity at 0.x 0 0 0 2 0 2 0 10 ( 10) 0 10 (10)(300) 2 6000 f x v f dv v a dx vdv dx v x v            





077.5 ft/s v  (b) Time to stop. 0 0 0 0 0 10 10 0 10 77.5 10 10 f t v f f dv a dx dv dt v t v t           





7.75 s f t  

The acceleration of a particle is defined by the relation _{a 3e}0.2t_{, where a and t are expressed in ft/s and} 2

seconds, respectively. Knowing that x 0 and v 0 at t 0, determine the velocity and position of the particle when t 0.5 s. SOLUTION Acceleration: _{a }3_e0.2t ft/s2 Given : v₀ 0 ft/s, x₀  0 ft Velocity: a dv dv adt dt   





0 0 0.2 0.2 0 0 0.2 3 0= 15 15 1 ft/s v t v t _t t t o t dv adt v v e dt v e v e           







Position: v dx dx vdt dt   













0 0 0.2 0.2 0 0 0.2 15 1 x 0=15 5 15 5 75 ft x t x t _t t t o t dx vdt x x e dt t e x t e             







Velocity at t=0.5 s _{15 1}



0.2 0.5



_ft/s v e 

 

0.5

1.427 ft/s

v



 Position at t=0.5 s _{15 0.5 5}



0.2 0.5



_{75 ft} x  e  

 

0.5

0.363 ft.

x





PROBLEM 11.11

The acceleration of a particle is directly proportional to the square of the time t. When t  the particle is 0, at x24 m. Knowing that at t6 s,x96 m and v18 m/s, express x and v in terms of t.

SOLUTION We have akt2 kconstant Now dv a kt2 dt   At t6 s,v18 m/s: 2 18 6 v t dv kt dt





or ₁₈ 1 ₍3 ₂₁₆₎ 3 v  k t  or ₁₈ 1 ₍ 3 _216)(m/s) 3 v  k t  Also ₁₈ 1 ₍ 3 ₂₁₆₎ 3 dx v k t dt     At t0,x24 m: 3 24 0 1 18 ( 216) 3 x t dx _  k t  _dt  





or 24 18 1 1 4 216 3 4 x  t k_ t  t_   Now At t6 s,x96 m: 96 24 18(6) 1 1(6)4 216(6) 3k 4      _  _   or 1 _m/s4 9 k Then _{24 18} 1 1 1 4 ₂₁₆ 3 9 4 x  t  _{ } t  t_    or _{( )} 1 4 _{10 24} 108 x t  t  t  and 18 1 1 (3 216) 3 9 v   _{ } t    or ( ) 1 3 10 27 v t  t  

The acceleration of a particle is defined by the relation _akt2_{. (a) Knowing that v 8 m/s when t  0} and that v 8 m/s when t2 s,determine the constant k. (b) Write the equations of motion, knowing also

that x when 0 t2 s. SOLUTION 2 2 a kt dv a kt dt    (1) 0, 8 m/s    t v and t2 s, v 8 ft/s (a) 8 2 2 8dv 0kt dt  





_{8 ( 8)} 1 ₍₂₎3 3    k _{6.00 m/s}4 k  (b) Substituting k6 m/s into (1)4 2 6   dv a t dt 2 6  a t  0, 8 m/s:    t v 2 8 06 v t dv t dt  





3 1 ( 8) 6( ) 3    v t v2t38  3 2 8    dx v t dt 2 s, 0:   t x 3 4 0 2 2 1 (2 8) ; 8 2 t x t dx t  dt x t  t





4 4 4 1 1 8 (2) 8(2) 2 2 1 8 8 16 2     _  _{ }  _         x t t x t t 1 4 8 8 2    x t t 

PROBLEM 11.13

A Scotch yoke is a mechanism that transforms the circular motion of a crank into the reciprocating motion of a shaft (or vice versa). It has been used in a number of different internal combustion engines and in control valves. In the Scotch yoke shown, the acceleration of Point A is defined by the relation a 1.8sin kt, where a and t are expressed in m/s2_{and seconds,}

respectively, and k 3 rad/s. Knowing that x 0 and v 0.6 m/s when t 0, determine the velocity and position of Point A when t 0.5 s.

SOLUTION Acceleration: a  1.8sin m/skt 2 Given : v₀ 0.6 m/s, x₀  0, k 3 rad/s Velocity: 0 0 v t v dv a dv adt dv adt dt    







0 _{0 0} 0 1.8 1.8 sin cos t t t v v a dt kt dt kt k  

_

 

_

 1.8 0.6 (cos 1) 0.6cos 0.6 3      v kt kt 0.6cos m/s v  kt Position: 0 0 x t x dx v dx vdt dx vdt dt    







0 0 0 ₀ 0.6 0.6 cos sin t t t x x v dt kt dt kt k  

_



_

 0.6 0 (sin 0) 0.2sin 3     x kt kt 0.2sin m x  kt When t =0.5 s, kt  (3)(0.5) 1.5 rad 0.6cos1.5 0.0424 m/s v   v  42.4 mm/s  0.2sin1.5 0.1995 m x   x 199.5 mm 

For the scotch yoke mechanism shown, the acceleration of Point A is defined by the relationa  1.08sinkt 1.44 cos ,kt where a and t are expressed in m/s2 and

seconds, respectively, and k = 3 rad/s. Knowing that x = 0.16 m and v = 0.36 m/s when t = 0, determine the velocity and position of Point A when t = 0.5 s.

SOLUTION

Acceleration: a  1.08sinkt1.44cos m/skt 2

Given : v₀ 0.36 m/s, x₀  0.16, k 3 rad/s Velocity: 0 0 v t v dv a dv adt dv adt dt    







Integrate: _{0 0 0} 0 0 1.08 sin 1.44 cos 1.08 1.44 0.36 cos sin 1.08_{(cos3 1)} 1.44_{(sin 3 0)} 3 3 0.36cos3 0.36 0.48sin 3 0.36cos3 0.48sin 3 m/s t t t t v v kt dt kt dt v kt kt k k t t t t v t t                





Evaluate at t 0.5 s 0.36cos1.50.360.48sin1.5 v  453 mm/s 

Position: 0 0 x t x dx v dx vdt dx vdt dt    







0 _{0 0 0} 0 0 0.36 cos 0.48 sin 0.36 0.48 0.16 sin cos 0.36 0.48 (sin 3 0) (cos3 1) 3 3 0.12sin 3 0.16cos3 0.16 0.12sin 3 0.16cos3 m t t t t t x x v dt kt dt kt dt x kt kt k k t t t t x t t                







PROBLEM 11.15

A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After contact the equipment experiences an acceleration of a kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.

SOLUTION

vdv

a k x

dx

  

Separate and integrate.

0 0 2 2 2 2 0 0 1 1 1 1 2 2 2 2 f f f v x v x f f vdv k x dx v v kx k x       





Use v₀ 4 m/s,x_f 0.02 m, andv_f  Solve for k. 0.

2 2 2 1 1 0 (4) (0.02) 40,000 s 2 2     k k Maximum acceleration. 2 max max: ( 40,000)(0.02) 800 m/s a  kx    2 800 m/s a 

A projectile enters a resisting medium at x with an initial velocity 0

v0900 ft/s and travels 4 in. before coming to rest. Assuming that the velocity of the projectile is defined by the relation vv₀ kx,where v is expressed in ft/s and x is in feet, determine (a) the initial acceleration of the projectile, (b) the time required for the projectile to penetrate 3.9 in. into the resisting medium. SOLUTION First note When 4 ft, 0: 12 x v 0 (900 ft/s) 4 ft 12 k   _{  }   or 27001 s k (a) We have vv₀kx Then a dv d (v₀ kx) kv dt dt      or a k v( ₀kx) At t 0: 2700 (900 ft/s 0)1 s a  or a₀  2.43 10 ft/s 6 2  (b) We have dx v v₀ kx dt    At t0,x 0: 0 ₀ 0 x _dx t dt v kx





or 1[ln( ₀ )]₀x v kx t k    or 0 0 0 1 1 1 ln ln 1 k v v t k v kx k x     _{ }   _  _{  }   _{ } When x3.9 in.: _{1 2700 1/s 3.9}





900 ft/s 12 s 1 1 ln 1 ft 2700 t _ _    or t1.366 10 s 3 

PROBLEM 11.17

The acceleration of a particle is defined by the relation a k x/ . It has been experimentally determined that 15 ft/s

v when x0.6 ft and that v9 ft/s when x1.2 ft. Determine (a) the velocity of the particle

when x1.5 ft, (b) the position of the particle at which its velocity is zero.

SOLUTION vdv k a dx x   

Separate and integrate using x0.6 ft,v15 ft/s.

15 0.6 2 15 _0.6 2 2 1 _ln 2 1 1 (15) ln 2 2 0.6 v x x v dx vdv k x v k x x v k          _{ }  





(1) When v9 ft/s, x1.2 ft 2 2 1 1 1.2 (9) (15) ln 2 2 k 0.6      _{ }   Solve for k. 2 2 103.874 ft /s k

(a) Velocity when 65 ft.x

Substitute _{k103.874 ft /s}2 2 _{and x1.5 ft into (1).}

2 2 1 1 1.5 (15) 103.874 ln 2v 2 0.6      _{ }   5.89 ft/s v 

(b) Position when for v 0,

2 1 0 (15) 103.874 ln 2 0.6 ln 1.083 0.6 x x      _{ }    _{ }       x1.772 ft 

A brass (nonmagnetic) block A and a steel magnet B are in equilibrium in a brass tube under the magnetic repelling force of another steel magnet C located at a distance x  0.004 m from B. The force is inversely proportional to the square of the distance between B and C. If block A is suddenly removed, the acceleration of block B is a 9.81k x/ ,2 where a and x are expressed in m/s2 and m, respectively, and _{k 4 10 m /s .}4 3 2 _{Determine the maximum velocity and} acceleration of B.

SOLUTION

The maximum velocity occurs when a0.

0 9.81 ₂

m k x    4 2 4 10 _{40.775 10 m}6 2 _{0.0063855 m} 9.81 9.81 m m k x x        

The acceleration is given as a function of x.

2 9.81     dv k v a dx x

Separate variables and integrate:

0 0 2 2 0 2 0 0 9.81 9.81 1 1 1 9.81( ) 2 v x x x x k dx vdv dx x dx vdv dx k x v x x k x x             _  _  







2 0 0 4 2 2 1 1 1 9.81( ) 2 1 1 9.81(0.0063855 0.004) (4 10 ) 0.0063855 0.004 0.023402 0.037358 0.013956 m /s m m m v x x k x x                    _  _       Maximum velocity: vm 0.1671 m/s vm167.1 mm/s 

The maximum acceleration occurs when x is smallest, that is, x0.004 m. 4 2 4 10 9.81 (0.004) m a      _{15.19 m/s}2 m a 

PROBLEM 11.19

Based on experimental observations, the acceleration of a particle is defined by the relation a (0.1 sin x/b), where a and x are expressed in m/s2 and meters, respectively. Knowing that b0.8 m and that v1 m/s

when x determine (a) the velocity of the particle when 0, x 1 m, (b) the position where the velocity is

maximum, (c) the maximum velocity.

SOLUTION We have 0.1 sin 0.8 dv x v a dx     _  _   When x0,v1 m/s: 1 0 0.1 sin0.8 v x _x vdv _  _dx  





or 2 0 1 ( 1) 0.1 0.8 cos 2 0.8 x x v   _ x _   or 1 2 0.1 0.8 cos 0.3 2 0.8 x v   x 

(a) When 1 m:x  1 2 0.1( 1) 0.8 cos 1 0.3

2v 0.8



    

or v 0.323 m/s 

(b) When vv_max, a 0: 0.1 sin 0 0.8 x   _  _   or x 0.080134 m x 0.0801 m  (c) When 0.080 134 m:x  2 max 2 2 1 0.080 134 0.1( 0.080 134) 0.8 cos 0.3 2 0.8 0.504 m /s v        or v_max 1.004 m/s 

A spring AB is attached to a support at A and to a collar. The unstretched length of the spring is l. Knowing that the collar is released from rest at

x



x

₀ and has an acceleration defined by the relationa 100(xlx/ l2x2), determine the velocity of

the collar as it passes through Point C.

SOLUTION Since a is function of x, 2 2 100     _  _    dv lx a v x dx _{l x}

Separate variables and integrate:

0 0 0 2 2 100 f v v x lx vdv x dx l x     _  _   





0 0 2 2 2 2 2 0 1 1 100 2 2 2      _   _   f x x v v l l x 2 2 0 2 2 2 0 1 0 100 2 2      _    _   f x v l l l x 2 2 2 2 2 2 0 0 2 2 2 0 1 100 ( 2 ) 2 2 100 ( ) 2 f v l x l l l x l x l          2 2 0 10( ) f v  l x   l

PROBLEM 11.21

The acceleration of a particle is defined by the relation



1 x



,

a  k e where k is a constant. Knowing that

the velocity of the particle is v   m/s when 9 x  m and that the particle comes to rest at the origin, 3

determine (a) the value of k, (b) the velocity of the particle when x  m. 2

SOLUTION Acceleration: _{a  k}



1_ex



Given : at x  m, 3 v  m/s 9 at x m, 0 v  m/s 0





k 1 x adx vdv e dx vdv   

Integrate using x=-3 m and v=9 m/s as the lower limits of the integrals









3 9 2 9 3 k 1 1 2 x v x x _v x e dx vdv k x e v        





Velocity:





₃ 3





1 2 1₍₉₎2 2 2 x k xe   e  v  (1)

(a) Now substitute v=0 m/s and x=0 m into (1) and solve for k







₀ 0 ₃ 3



1₀2 1₍₉₎2 2 2 k e   e   2 2 2.52 m /s k 

(b) Find velocity when x= -2 m using the equation (1) and the value of k







2 3



1 2 1 2 2.518 2 3 (9) 2 2 e e v        4.70 m/s v 

Starting from x  0 with no initial velocity, a particle is given an acceleration _0.1 2 _16,

a v  where a and v are expressed in ft/s2 and ft/s, respectively. Determine (a) the position of the particle when v  3ft/s, (b) the speed and acceleration of the particle when x  4 ft.

SOLUTION 2 1/2 0.1( 16) vdv a v dx    (1)

Separate and integrate.

2 0 ₁₆ 00.1 v _vdv x dx v  





2 1/2 0 2 1/2 2 1/2 ( 16) 0.1 ( 16) 4 0.1 10[( 16) 4] v v x v x x v         (2) (a) v3 ft/s. x10[(3216)1/2  4] x10.00 ft  (b) x4 ft. From (2), (v216)1/2  4 0.1x 4 (0.1)(4) 4.4 2 2 2 2 16 19.36 3.36ft /s v v    v1.833 ft/s  From (1), a0.1(1.833216)1/2 a0.440 ft/s2 

PROBLEM 11.23

A ball is dropped from a boat so that it strikes the surface of a lake with a speed of 16.5 ft/s. While in the water the ball experiences an acceleration of a10 0.8 , v _{where a and v}

are expressed in ft/s2 and ft/s, respectively. Knowing the ball takes 3 s to reach the bottom of the lake, determine (a) the depth of the lake, (b) the speed of the ball when it hits the bottom of the lake.

SOLUTION 10 0.8 dv a v dt   

Separate and integrate:

010 0.8 0 v t v dv dt v 





0 0 1 ln(10 0.8 ) 0.8 10 0.8 ln 0.8 10 0.8 v v v t v t v      _{ }  _    0.8 0 10 0.8 (10 0.8 ) t v v e    or ₀ 0.8 0.8 0 0.8 10 (10 0.8 ) 12.5 (12.5 ) t t v v e v v e         With v₀ 16.5ft/s 12.5 4 0.8t v  e

Integrate to determine x as a function of t.

dx 12.5 4 0.8t v e dt     0.8 0 0(12.5 4 ) x t t dx  e dt





0.8 0.8 0 12.5 5 tt 12.5 5 t 5 x t e  t e  (a) At 35 s,t 2.4 12.5(3) 5 5 42.046 ft x  e   x42.0 ft  (b) _{v12.5 4 e}2.4_{12.863 ft/s 12.86 ft/s} v 

PROBLEM 11.24

The acceleration of a particle is defined by the relation a k v,where k is a constant. Knowing that x  0 and v  81 m/s at t  0 and that v  36 m/s when x  18 m, determine (a) the velocity of the particle when

x  20 m, (b) the time required for the particle to come to rest.

SOLUTION (a) We have vdv a k v dx    so that v dv k dx When x0,v81 m/s: 81 0 v x v dv k dx





or 2[ 3/2]₈₁ 3 v v   kx or 2[ 3/2 729] 3 v    kx When x18 m,v36 m/s: 2(363/2 729) (18) 3   k or k19 m/s2 Finally When x20 m: 2( 3/2 729) 19(20) 3 v    or v3/2159 v29.3 m/s  (b) We have dv a 19 v dt    At t0,v81 m/s: 81 0 19 v _dv t dt v  





or 2[ _v]₈₁v _{ }19_t or 2( v  9) 19t When v 0: 2( 9)  19t or t0.947 s 

The acceleration of a particle is defined by the relation a  kv2.5, where k is a constant. The particle starts at 0

x  with a velocity of 16 mm/s, and when x  mm the velocity is observed to be 4 mm/s. Determine 6

(a) the velocity of the particle when x  mm, (b) the time at which the velocity of the particle is 9 mm/s. 5

SOLUTION Acceleration: a  kv2.5 Given : at t=0, x  mm, 0 v 16mm/s at x mm, 6 v  mm/s 4 2.5 adx vdv kv dx vdv    Separate variables _kdxv3 2_dv

Integrate using x=-0 m and v=16 mm/s as the lower limits of the integrals 3 2 0 16 1 2 16 0 -k 2 x v x v dx v dv kx v      





Velocity and position: ₂ 1 2 1 2

kx v  (1)

Now substitute v=4 mm/s and x=6 mm into (1) and solve for k

 

1 2 3 2 1 2 1 6 4 2 0.0833 mm k k s     

(a) Find velocity when x= 5 mm using the equation (1) and the value of k

 

_{5 2} 1 2 1 2 k  v  4.76 mm/s v  (b) a dv or adt dv dt   or dv a adt dv dt   Separate variables _kdtv2.5_dv

PROBLEM 11.25 (Continued)

Integrate using t=0 and v=16 mm/s as the lower limits of the integrals 5 2 0 16 3 2 16 0 -k 2 3 t v t _v dt v dv kt v      





Velocity and time: 2 3 2 1

3 96

kt v  (2)

Find time when v= 9 mm/s using the equation (2) and the value of k

 

3 2 2 1 9 3 96 kt   0.171 s t 

A human powered vehicle (HPV) team wants to model the acceleration during the 260 m sprint race (the first 60 m is called a flying start) using a = A – Cv2, where a is acceleration in m/s2 and v is the velocity in m/s. From wind tunnel testing, they found that C = 0.0012 m-1. Knowing that the cyclist is going 100 km/h at the 260 meter mark, what is the value of A?

SOLUTION Acceleration: a  A Cv2m/s2 Given : C  0.0012 m1, v_f 100 km/hr when x_f 260 m Note: 100 km/hr = 27.78 m/s



2



adx vdv A Cv dx vdv    Separate variables dx vdv ₂ A Cv  

Integrate starting from rest and traveling a distance xf with a final velocity vf.









 

2 0 0 2 0 0 2 2 1 ln 2 1 1 ln ln 2 2 1 ln 2 f f f f x v v x f f f f vdv dx A Cv x A Cv C x A v A C C A x C A Cv               _{ }   





Next solve for A

2 2 2 1 f f Cx f Cx Cv e A e   

PROBLEM 11.27

Experimental data indicate that in a region downstream of a given louvered supply vent the velocity of the emitted air is defined by

0 0.18 / ,

v v x where v and x are expressed in m/s and meters,

respectively, and v₀ is the initial discharge velocity of the air. For 0 3.6 m/s,

v  determine (a) the acceleration of the air at x2 m,

(b) the time required for the air to flow from x to x 3 m.1 

SOLUTION (a) We have 0 0 2 0 3 0.18 0.18 0.0324 dv a v dx v d v x dx x v x     _{ }     When x2 m: 2 3 0.0324(3.6) (2) a  or _{a 0.0525 m/s}2 _ (b) We have dx _v 0.18v0 dt   x From x1 m tox3 m: 3 1 3 0 1 0.18 t t xdx v dt





or 3 2 0 3 1 1 1 0.18 ( ) 2x v t t  _{  }     or 12 3 1 (9 1) ( ) 0.18(3.6) t t   or t3 t1 6.17 s 

Based on observations, the speed of a jogger can be approximated by the relation 0.3

7.5(1 0.04 ) ,

v  x where v and x are expressed in mi/h and miles, respectively.

Knowing that x at 0 t determine (a) the distance the jogger has run when 0, t1 h,

(b) the jogger’s acceleration in ft/s2 at t 0, (c) the time required for the jogger to run 6 mi.

SOLUTION (a) We have dx v 7.5(1 0.04 )x 0.3 dt    At t0, x 0: _0.3 0 _{(1 0.04 )} 07.5 x _dx t dt x  





or 0.7 0 1 1 [(1 0.04 ) ] 7.5 0.7 0.04 x x t _  _{ }     or 1 (1 0.04 )  x 0.7 0.21t (1) or 1 _{[1 (1 0.21 )}1/0.7_] 0.04 x   t At t1 h: 1 {1 [1 0.21(1)]1/0.7} 0.04 x   or x7.15 mi  (b) We have 0.3 0.3 2 0.3 0.7 0.4 7.5(1 0.04 ) [7.5(1 0.04 ) ] 7.5 (1 0.04 ) [(0.3)( 0.04)(1 0.04 ) ] 0.675(1 0.04 ) dv a v dx d x x dx x x x              At t0, x 0: 2 2 0 5280 ft 1 h 0.675 mi/h 1 mi 3600 s a      _   or a0  275 10 ft/s 6 2  (c) From Eq. (1) 1 [1 (1 0.04 ) ]0.7 0.21 t   x When x6 mi: 1 {1 [1 0.04(6)] }0.7 0.21 0.83229 h t    or t49.9 min 

PROBLEM 11.29

The acceleration due to gravity at an altitude y above the surface of the earth can be expressed as 6 2 32.2 [1 ( /20.9 10 )] a y    

where a and y are expressed in ft/s2 and feet, respectively. Using this expression, compute the height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is (a) 1800 ft/s, (b) 3000 ft/s, (c) 36,700 ft/s.

SOLUTION We have



6



2 20.9 10 32.2 1 y dv v a dy      When y0, v v₀

provided that v does reduce to zero, yymax, v 0 Then





max 0 6 0 2 0 20.9 10 32.2 1 y v _y v dv dy    





or max 6 2 6 0 20.9 10 ₀ 1 1 32.2 20.9 10 2 1 y y v            _    or max 6 2 6 0 20.9 10 1 1345.96 10 1 1 y v        _  _      or ₂ 0 6 2 0 max 20.9 10 64.4 v v y    (a) v₀ 1800 ft/s: ₂ 6 2 max ₍₁₈₀₀₎ 20.9 10 (1800) 64.4 y    or y_max 50.4 10 ft 3  (b) v₀ 3000 ft/s: 2 6 2 max ₍₃₀₀₀₎ 20.9 10 (3000) 64.4 y    or y_max 140.7 10 ft 3 

(c) v₀ 36,700 ft/s: 2 6 2 10 max _(36,700) 20.9 10 (36,700) 3.03 10 ft 64.4 y      

This solution is invalid since the velocity does not reduce to zero. The velocity 36,700 ft/s is above the escape velocity v_R from the earth. For v_R and above.

max

PROBLEM 11.30

The acceleration due to gravity of a particle falling toward the earth is a gR r2/ ,2 where r is the distance from the center of the earth to the particle, R is the radius of the earth, and g is the acceleration due to gravity at the surface of the earth. If R3960 mi,calculate the

escape velocity, that is, the minimum velocity with which a particle must be projected

vertically upward from the surface of the earth if it is not to return to the earth. (Hint: 0 v for r  .) SOLUTION We have 2 2 dv gR v a dr    r When , , 0 e r R v v r v      then 0 ₂2 e v R gR vdv dr r   





or 1 2 2 1 2 e R v gR r      _{ }   or 1/2 2 2 5280 ft 2 32.2 ft/s 3960 mi 1 mi e v  gR   _    _   or _{36.7 10 ft/s}3 e v   

The velocity of a particle is vv₀[1 sin ( / )]. t T Knowing that the particle starts from the origin with an

initial velocity v₀, determine (a) its position and its acceleration at t3 ,T (b) its average velocity during the

interval t to 0 t T.

SOLUTION

(a) We have dx v v₀ 1 sin t

dt T       _  _{ }     At t0,x 0: ₀ 0 0 1 sin x t _t dx v dt T      _  _{ }    





0 0 0 cos cos t T t T t T x v t v t T T              _  _{ }  _  _{ } _         (1) At t3 :T x_3T v₀ 3T Tcos 3T T v₀ 3T 2T T             _  _{ } _ _  _       x3T 2.36v T0  0 1 sin 0 cos dv d t t a v v dt dt T T T             _{ }  _{ }         At t3 :T a_3T v₀ cos 3T T T     0 3T v a T    (b) Using Eq. (1) At t 0: x₀ v₀ 0 Tcos(0) T 0      _   _   At :tT x_T v₀ T T cos T T v₀ T 2T 0.363v T₀ T            _  _{ } _ _  _       Now 0 0 ave 0.363 0 0 T x x v T v t T       vave 0.363v0 

Problem 11.32

An eccentric circular cam, which serves a similar function as the Scotch yoke mechanism in Problem 11.13, is used in conjunction with a flat face follower to control motion in pumps and in steam engine valves. Knowing that the eccentricity is denoted by e, the maximum range of the displacement of the follower is dmax, and the maximum velocity of the follower is vmax, determine the displacement, velocity, and acceleration of the follower.

SOLUTION

Constraint: y  r ecos (1)

Differentiate: y   esin (2)

Differentiate again: y e



sin



e



2cos



(3)

ymax occurs when cosθ =1 and ymin occurs when cosθ = -1





max max min max max 2 d y y d r e r e d e        max 2 d e  (4)

Substitute (4) into (1) to get Position max _cos

2

d

y r  

Max Velocity occurs when

sin



 

1

max v _ e max v e    (5)

Substitute (5) into (2) to get velocity and assume cw rotation. y  v_maxsin Substitute (4) and (5) into (3)

2

max max max

2 max 4 sin cos 2 2 d d v y d         _ _     Acceleration: 2 max max max 2 sin cos 2 d v y d         

An airplane begins its take-off run at A with zero velocity and a constant acceleration a. Knowing that it becomes airborne 30 s later at B and that the distance AB is 900 m, determine (a) the acceleration

a, (b) the take-off velocity

v

_B.

SOLUTION Given : x_A  , 0 x_B 900 m,v_A  , 0 t 30 s Uniform Acceleration: 1 2 2 B A A x  x v t at

Substitute known values: 900 m 0 0 1



30 s



2 2a   

(a) Solve for acceleration a2.0 m/s2

Uniform Acceleration: v_B v_Aat

Substitute known values: v_B  0



2 m/s2





30 s



PROBLEM 11.34

A motorist is traveling at 54 km/h when she observes that a traffic light 240 m ahead of her turns red. The traffic light is timed to stay red for 24 s. If the motorist wishes to pass the light without stopping just as it turns green again, determine (a) the required uniform deceleration of the car, (b) the speed of the car as it passes the light.

SOLUTION

Uniformly accelerated motion:

0 0 0 54 km/h 15 m/s x  v   (a) 2 0 0 1 2 xx v t a t when t24s, x240 m: 2 2 1 240 m 0 (15 m/s)(24 s) (24 s) 2 0.4167 m/s a a      _{0.417 m/s}2 a   (b) vv₀a t when t24s: (15 m/s) ( 0.4167 m/s)(24 s) 5.00 m/s 18.00 km/h v v v      v18.00 km/h 

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a 750-ft ramp at a high speed v₀ and travels 540 ft in 6 s

at constant deceleration before its speed is reduced to v₀/ 2. Assuming the same constant deceleration, determine (a) the additional time required for the truck to stop, (b) the additional distance traveled by the truck.

SOLUTION Given : x_o  , 0 x_A 540 m, t_A  6 s, 1 2 A o v  v , L_ramp 750ft Uniform Acceleration: v_A v_o at_A

Substitute known values: 1

  

6 2vo vo  a 1 12 o a   v Uniform Acceleration: 1 2 2 A o o A A x  x v t  at

Substitute known values: 540 0

 

6 1

 

6 2 24 o o v v    120 ft/s o v  and a  10 ft/s2 (a) v_B v_o at_B

Substitute known values: 0 120 10 t_B

Solve for tB tB 12 s

Additional time to stop t_B  t_A 6.0 s 

(b) 1 2

2

B o o B B

x  x v t  at

Substitute known values: 0 120 12

 

110 12

 

2 2

B

x   

Solve for xB xB 720 ft

PROBLEM 11.36

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 16 s later. Knowing that the descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude and assuming that g32.2 ft/s ,2 determine (a) the speed v₁of the

rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.

SOLUTION (a) We have _{1 1} 1 2 2 yy v t at At t_land, y 0 Then ₁ 2 2 0 89.6 ft (16 s) 1 ( 32.2 ft/s )(16 s) 2 v     or v₁252 ft/s  (b) We have v2v₁22 (a yy₁) At y y_max, v0 Then 2 2 max 0 (252 ft/s)  2( 32.2 ft/s )(y 89.6) ft or y_max1076 ft 

A small package is released from rest at A and moves along the skate wheel conveyor ABCD. The package has a uniform acceleration of 4.8

2

m/s as it moves down sections AB and CD, and its velocity is constant between B and C. If the velocity of the package at D is 7.2 m/s, determine (a) the distance d between C and D, (b) the time required for the package to reach D.

SOLUTION

(a) For A B and C D we have 2 2 0 2 ( 0) v v  a xx Then, at B 2 2 2 2 0 2(4.8 m/s )(3 0) m 28.8 m /s ( 5.3666 m/s) BC BC v v      and at D v_D2 v_BC2 2a_CD(x_Dx_C) d x_D x_C or (7.2 m/s)2 (28.8 m /s ) 2(4.8 m/s )d2 2  2 or d2.40 m 

(b) For A B and C D we have

0 vv at Then A B 5.3666 m/s 0 (4.8 m/s )  2 t_AB or t_AB 1.118 04 s and C D 7.2 m/s 5.3666 m/s (4.8 m/s )  2 t_CD or t_CD 0.38196 s

Now, for B C, we have x_C x_B v_{BC BC}t

or 3 m(5.3666 m/s)t_BC

or t_BC 0.55901 s

Finally, t_D t_AB t_BC t_CD (1.11804 0.55901 0.38196) s 

PROBLEM 11.38

A sprinter in a 100-m race accelerates uniformly for the first 35 m and then runs with constant velocity. If the sprinter’s time for the first 35 m is 5.4 s, determine (a) his acceleration, (b) his final velocity, (c) his time for the race.

SOLUTION Given: 0 x 35 m, aconstant 35 m x 100 m, vconstant At t0, v0 when x35 m, t5.4 s Find: (a) a (b) v when x100 m (c) t when x100 m (a) We have _{0 0} 1 2 2 x  t at for 0 x 35 m At t5.4 s: 35 m 1 (5.4 s)2 2a  or _{a2.4005 m/s}2 2 2.40 m/s a 

(b) First note that vv_maxfor 35 m x 100 m.

Now v2  0 2 (a x0) for 0 x 35 m When x35 m: v_max2 2(2.4005 m/s )(35 m)2 or v_max 12.9628 m/s v_max 12.96 m/s  (c) We have xx₁v t₀( t₁) for 35 m x 100 m When x100 m: 100 m35 m (12.9628 m/s)( t₂5.4) s or t₂10.41 s 

Automobile A starts from O and accelerates at the constant rate of 0.75 m/s2. A short time later it is passed by bus B which is traveling in the opposite direction at a constant speed of 6 m/s. Knowing that bus B passes point O 20 s after automobile A started from there, determine when and where the vehicles passed each other.

SOLUTION Place origin at 0. Motion of auto.

 

x_{A 0} 0,

 

v_{A 0}0, a_A0.75 m/s2

   

2





2 0 0 1 1 0 0 0.75 2 2 A A A A x  x  v t  a t       t   2 0.375 m A x  t Motion of bus.

 

xB 0?,

 

vB 0  6 m/s, aB  0

   

_{0 0}

 

₀ 6 m B B B B x  x  v t  x  t At t 20 ,s x_B 0.

    

₀ 0 x_B  6 20

 

x_{B 0}120 m Hence, x_B 120 6t

When the vehicles pass each other, x_B  x_A. 2 1206t 0.375t 2 0.375t 6t 120 0

 





 



2 6 (6) 4 0.375 120 2 0.375 t      6 14.697 11.596 s and 27.6 s 0.75 t     

Reject the negative root. t 11.60 s

Corresponding values of xA and xB.







2 0.375 11.596 50.4 m A x  

 



120 6 11.596 50.4 m B x    x50.4 m

PROBLEM 11.40

In a boat race, boat A is leading boat B by 50 m and both boats are traveling at a constant speed of 180 km/h. At t the boats 0, accelerate at constant rates. Knowing that when B passes A, t 8 s and v_A 225 km/h, determine (a) the acceleration of A, (b) the acceleration of B. SOLUTION (a) We have v_A ( )v_{A 0}a t_A 0 ( )v_A 180 km/h 50 m/s At

t



8 s:

vA 225 km/h 62.5 m/s Then 62.5 m/s 50 m/s a_A(8 s) or

a

_A



1.563 m/s

2  (b) We have 2 2 2 0 0

1

1

( )

( )

50 m (50 m/s)(8 s)

(1.5625 m/s )(8 s)

500 m

2

2

A A A A

x



x



v

t



a t









and

0 ( )

₀

1

2

2

B B B

x

 

v

t



a t

( )vB ₀ 50 m/s At

t



8 s:

x_A x_B

500 m (50 m/s)(8 s)

1

(8 s)

2

2

a

B





or

a

_B



3.13 m/s

2 

As relay runner A enters the 65-ft-long exchange zone with a speed of 30 ft/s, he begins to slow down. He hands the baton to runner B 2.5 s later as they leave the exchange zone with the same velocity. Determine (a) the uniform acceleration of each of the runners, (b) when runner B should begin to run.

SOLUTION

Let x  at the start of the exchange zone, and 0 t  0 as runner A enters the exchange zone.

Then,

 

vA 0  30 ft/s. Motion of runner : _A

 

0 A A A v  v  a t

 

2 0 1 2 A A A x  v t  a t

(a) Solving for a_A, and noting that x_A 65 ft when t 2.5 s

 

₀

  

2 2 2 2 65 30 2.5 2.5 A A A x v t a t             _{3.20 ft/s} 2 A a    At 2.5 t  s,

 

v_{A f} 30 



3.20 2.5

 

 22.0 ft/s Motion of runner B:

 

vB 0  0 and xB  0 at the starting time t_B of runner B.

Then, v2B 

 

vB 02  2a xB B

Solving for ,a_B and noting that v_B 

 

v_{A f}  22.0 ft/s

When 65 x_B  ft,

   

  

2 2 ₂ 0 22.0 0 2 2 65 B B B B v v a x     _{3.723 ft/s} 2 B a  

 

₀









B B B B B B v  v  a t t  a t t B B B v t t a  

(b) Solving for ,t_B and noting that v_B 22.0 ft/s when t 2.5 s

22.0 2.5 3.41 s 3.723 B B B v t t a      

PROBLEM 11.42

Automobiles A and B are traveling in adjacent highway lanes and at t  have the positions and speeds shown. Knowing 0 that automobile A has a constant acceleration of _{1.8 ft/s}2 _and that B has a constant deceleration of 1.2 ft/s , determine 2 (a) when and where A will overtake B, (b) the speed of each automobile at that time.

SOLUTION 2 2 0 0 1.8 ft/s 1.2 ft/s ( ) 24 mi/h 35.2 ft/s ( ) 36 mi/h 52.8 ft/s A B A B a a v v         Motion of auto A: 0 ( ) 35.2 1.8 A A A v  v a t   t (1) 2 2 0 0 1 1 ( ) ( ) 0 35.2 (1.8) 2 2 A A A A x  x  v t a t   t t (2) Motion of auto B: 0 ( ) 52.8 1.2 B B B v  v a t   t (3) 2 2 0 0 1 1 ( ) ( ) 75 52.8 ( 1.2) 2 2 B B B B x  x  v t a t   t  t (4) (a) A overtakes at B t t₁. 2 2 1 1 1 : 35.2 0.9 75 52.8 0.6 A B x x t t   t  t 2 1 1 1.5t 17.6t 75 0 1 3.22 s and 1 15.0546 t   t  t₁15.05 s  Eq. (2): x_A 35.2(15.05) 0.9(15.05) 2 x_A734 ft  A overtakes B

(b) Velocities when t₁15.05 s Eq. (1): v_A35.2 1.8(15.05) 62.29 ft/s A v  v_A42.5 mi/h  Eq. (3): v_B 52.8 1.2(15.05) 34.74 ft/s B v  v_B 23.7 mi/h 

PROBLEM 11.43

Two automobiles A and B are approaching each other in adjacent highway lanes. At t A and B are 3200 ft 0, apart, their speeds are v_A 65 mi/h and v_B 40 mi/h, and they are at Points P and Q, respectively. Knowing

that A passes Point Q 40 s after B was there and that B passes Point P 42 s after A was there, determine (a) the uniform accelerations of A and B, (b) when the vehicles pass each other, (c) the speed of B at that time.

SOLUTION (a) We have 2 0 0 1 0 ( ) ( ) 65 mi/h 95.33 ft/s 2 A A A A x   v t a t v   (x is positive ; origin at P.) At t40 s: 3200 m (95.333 m/s)(40 s) 1 (40 s)2 2aA   _{0.767 ft/s}2 A a    Also, 2 0 1 0 ( ) 2 B B B x   v t a t ( )v_{B 0} 40 mi/h58.667 ft/s (xB is positive ; origin at Q.) At t42 s: 3200 ft (58.667 ft/s)(42 s) 1 (42 s)2 2aB   or a_B 0.83447 ft/s2 a_B 0.834 ft/s2 

(b) When the cars pass each other x_Ax_B 3200 ft

Then (95.333 ft/s) 1( 0.76667 ft/s)2 (58.667 ft/s) 1(0.83447 ft/s )2 2 3200 ft 2 2 AB AB AB AB t   t  t  t  or 0.03390t2_AB154t_AB3200 0 Solving t20.685 s and t 4563 s t 0 t_AB 20.7 s  (c) We have v_B ( )v_{B 0}a t_B At tt_AB: v_B 58.667 ft/s (0.83447 ft/s )(20.685 s) 2 75.927 ft/s  v_B 51.8 mi/h 