# Engineering Mechanics Dynamics 12th Edition

(Parte 5 de 6)

(b) Fig. 12-9

It �s v dt f 0

(b) Fig. 12-10 using a = dv/dt, written as

Llv = change in velocity

Hence, to construct the v-t graph, we begin with the particle's initial velocity Vo and then add to this small increments of area (Llv) determined from the a-t graph. In this manner successive points,

VI = Vo + Llv, etc., for the v-t graph are determined, Fig. 12-9b. Notice that an algebraic addition of the area increments of the a-t graph is necessary, since areas lying above the t axis correspond to an increase in

V ("positive" area), whereas those lying below the axis indicate a decrease in v ("negative" area).

Similarly, if the v-t graph is given, Fig. 12-10a, it is possible to determine the s-t graph using v = ds/dt, written as

Lls = J v dt area under displacement = v-t graph

In the same manner as stated above, we begin with the particle's initial position So and add (algebraically) to this small area increments Lls determined from the v-t graph, Fig. 12-10b.

If segments of the a-t graph can be described by a series of equations, then each of these equations can be integrated to yield equations describing the corresponding segments of the v-t graph. In a similar manner, the s-t graph can be obtained by integrating the equations which describe the segments of the v-t graph. As a result, if the a-t graph is linear (a first-degree curve), integration will yield a v-t graph that is parabolic (a second-degree curve) and an s-t graph that is cubic (third degree curve).

12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION

The v-s and a-s Graphs. If the a-s graph can be constructed, then points on the v-s graph can be determined by using v dv = ads.

Integrating this equation between the limits v = Vo at s = So and v = VI at s = SI, we have,

�(vi -v6) = iSl a ds area under a-s graph ao va

Sl v f' ads =-(v 2 -V 2) a 2 I a

(a) s

�-------------------s

Therefore, if the red area in Fig. 12-11a is determined, and the initial velocity Vo at So = 0 is known, then VI = (21's> ds + V6)I/Z, Fig. 12-11b.

Successive points on the v-s graph can be constructed in this manner. If the v-s graph is known, the acceleration a at any position s can be determined using a ds = v dv, written as a = v(�:)

velocity times acceleration = slope of v-s graph

Thus, at any point (s, v) in Fig. 12-12a, the slope dvjds of the v-s graph is measured. Then with v and dvjds known, the value of a can be calculated, Fig. 12-12b.

The v-s graph can also be constructed from the a-s graph, or vice versa, by approximating the known graph in various intervals with mathematical functions, v = f(s) or a = g(s), and then using ds = v dv to obtain the other graph.

Sj v (b)

Fig. 12-1 s I---s--I a = v(dv/ds)

L-----�----------s I--s---l

(b) Fig. 12-12

2 CHAPTER 12 KINEMATICS OF A PARTICLE EXAMPLE 12.6

v (ft/s) v = 2t v 20 20

10 30 (b)

 2f--- ,

A bicycle moves along a straight road such that its posItIon is described by the graph shown in Fig. 12-13a. Construct the v-t and a-t graphs for 0 ::; t ::; 30 s.

s (ft)

500 f----------

 ==----1-'--0-----3'-0-t (s)

(a) SOLUTION v-t Graph. Since v = dsldt, the v-t graph can be determined by differentiating the equations defining the s-t graph, Fig. 12-13a. We have ds o ::; t < 10 s; s = (t2) ft v = dt = (2t) ft/s ds 10 s < t ::; 30 s; s = (20t -100) ft v = -= 20 ft/s dt t (s) The results are plotted in Fig. 12-13b. We can also obtain specific values of v by measuring the slope of the s-t graph at a given instant. For example, at t = 20 s, the slope of the s-t graph is determined from the straight line from 10 s to 30 s, i.e., v = Lls = 500 ft -100 ft = 20 ft/s t = 20 s; Llt 30 s -10 s a-t Graph. Since a = dvl dt, the a-t graph can be determined by differentiating the equations defining the lines of the v-t graph. This yields o ::; t < 10 s; v = (2t) ftls

10 < t ::; 30 s; v = 20 ftls dv a = -= 2 ft/s2 dt dv a =-= 0

L----1-'-0-----3-0-t (s) dt The results are plotted in Fig. 12-13c.

Fig. 12-13

NOTE: Show that a = 2 ft/s2 when t = 5 s by measuring the slope of the v-t graph.

12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION

EXAMPLE 12.7

The car in Fig. 12-14a starts from rest and travels along a straight track such that it accelerates at 10 m/s2 for 10 s, and then decelerates at 2 m/s2. Draw the v-t and s-t graphs and determine the time t' needed to stop the car. How far has the car traveled?

SOLUTION v<-t Graph. Since dv = a dt, the v-t graph is determined by integrating the straight-line segments of the a-t graph. Using the initial condition v = 0 when t = 0, we have -2 t' f----1--,-0+---A-:-2----iI--{ (s) '-----"------'

o :5 t < 10 s; a = (10) m/s2; lv dv = lt 10 dt, v = lOt

When t = 10 s, v = 10(10) = 100 m/s. Using this as the initial condition for the next time period, we have

10 s < t :5 t'; a = (-2) m/s2; r dv = r -2 dt, v = (-2t + 120) m/s J100m/s JlOs ( /) When t = t' we require v = O. This yields, Fig. 12-14b, vms

t' = 60 A v = lOt ns. 100 A more direct solution for t' is possible by realizing that the area under the a-t graph is equal to the change in the car's velocity. We require �v = 0 = Ai + A2, Fig. 12-14a. Thus o = 10 m/s2(10 s) + (-2 m/s2)(t' -10 s) t' = 60 s Ans.

s-t Graph. Since ds = v dt, integrating the equations of the v-t graph yields the corresponding equations of the s-t graph. Using the initial condition s = 0 when t = 0, we have o :5 t :5 10 s; v = (lOt) m/s; ls ds = lt lOt dt, s = (5t2) m When t = 10 s, s = 5(10? = 500 m. Using this initial condition,

10 s :5 t :5 60 s; v = (-2t + 120) m/s; r ds = r (-2t + 120) dt Jsoo m JlOs

S -500 = -t2 + 120t -[-(10? + 120(10)] s = (-t2 + 120t -600) m When t' = 60 s, the position is s = -(60)2 + 120(60) -600 = 3000 m Ans.

s (m) 3000 t' = 60 t(s)

(b)

The s-t graph is shown in Fig. 12-14c. Io£---'----{ (s)

NOTE: A direct solution for s is possible when t' = 60 s, since the triangular area under the v-t graph would yield the displacement �s = s -0 from t = 0 to t' = 60 Hence,

�s = !(60 s)(100 m/s) = 3000 m Ans.

10 60 (c)

Fig. 12-14

24 CHAPTER 12 KINEMATICS OF A PARTICLE EXAMPLE 12.8

v (ft/s) v = O.Zs + 10 v = 50

a = 0.04s + Z a=O L----I---_---s (ft) ZOO 400 (b)

Fig. 12-15

The v-s graph describing the motion of a motorcycle is shown in Fig. 12-15a. Construct the a-s graph of the motion and determine the time needed for the motorcycle to reach the position s = 400 ft.

SOLUTION a-s Graph. Since the equations for segments of the v-s graph are given, the a-s graph can be determined using a ds = v dv.

o :::; s < 200 ft; v = (0.2s + 10) ft/s dv d a = v ds = (0.2s + 10) ds (0.2s + 10) = 0.04s + 2 200 ft < S :::; 400 ft; v = 50 ft/s dv d a = v-= (50)-(50) = 0 ds ds The results are plotted in Fig. 12-15b.

Time. The time can be obtained using the v-s graph and v = ds/dt, because this equation relates v, s, and t. For the first segment of motion, s = 0 when t = 0, so o s < 200 ft; v = (O.2s + 10) ft/s; it is ds dt - o 0 0.2s + 10 dt = ds = ds v 0.2s + 10 t = (5 In(0.2s + 10) -5 In 10) s

At s = 200 ft, t = 51n[0.2(200) + 10] -5 In 10 = 8.05 s. Therefore, using these initial conditions for the second segment of motion,

200 ft < s :::; 400 ft; v = 50 ft/s; it is ds dt = -. S.05 s 200 50 ' dt = ds = ds v 50 t -8.05 = 5s0 -4; t = (5s 0 + 4.05 ) s

Therefore, at s = 400ft,

400 t = 50 + 4.05 = 12.0 s Ans.

NOTE: The graphical results can be checked in part by calculating slopes.

For example, at s = 0, a = v(dv/ds) = 10(50 -10)/200 = 2 m/s2. Also, the results can be checked in part by inspection. The v-s graph indicates the initial increase in velocity (acceleration) followed by constant velocity (a = 0).

F12-9. The particle travels along a straight track such that its position is described by the s-( graph. Construct the v-( graph for the same time interval.

s (m)

108 s = 108 s = 0.5 f

 =-------+---+---+-1 (s)

F12-9

Fl2-10. A van travels along a straight road with a velocity described by the graph. Construct the s-t and a-( graphs during the same period. Take s = 0 when t = O.

v (ft/s)

80 v = -41 + 80

 L------�f -1 (s)

F12-10

Fl2-1. A bicycle travels along a straight road where its velocity is described by the v-s graph. Construct the a-s graph for the same time interval.

v (m/s)

�--------------4�0-----s( m) Fl2-1

12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 25

F12-12. The sports car travels along a straight road such that its position is described by the graph. Construct the v-( and a-( graphs for the time interval 0 :S ( :S 10 s.

s (m) s = 301 -75 s = 3? o --'-"""-------f-------+--( (s) 5 10

Fl2-12

F12-13. The dragster starts from rest and has an acceleration described by the graph. Construct the v-( graph for the time interval 0 :S t :S (', where (' is the time for the car to come to rest.

(o f-------5+---------+-1---( (s) -10

F12-13

F12-14. The dragster starts from rest and has a velocity described by the graph. Construct the s-( graph during the time interval O:s ( :s 15 s. Also, determine the total distance traveled during this time interval.

v (m/s) v = 301 150

,v = -151 + 225

F12-14

26 CHAPTER 12 KINEMATICS OF A PARTICLE

12-42. The speed of a train during the first minute has been recorded as follows:

t (s) v (m/s) o 20 o 16 40 60 21 24

Plot the v-t graph, approximating the curve as straight-line segments between the given points. Determine the total distance traveled.

12-43. A two-stage missile is fired vertically from rest with the acceleration shown. In 15 s the first stage A burns out and the second stage B ignites. Plot the v-t and s-t graphs which describe the two-stage motion of the missile for o � t � 20 s.

25 18+-------------�

'---------+---+---t (s) 15 20

Prob. 12-43

*12-4. A freight train starts from rest and travels with a constant acceleration of 0.5 ft/S2. After a time t' it maintains a constant speed so that when t = 160 s it has traveled 2000 ft. Determine the time t' and draw the v-t graph for the motion.

-12-45. If the position of a particle is defined by s = [2 sin (7T/5)t + 4] m, where t is in seconds, construct the s-t, v-t, and a-t graphs for 0 � t � 10 s.

12-46. A train starts from station A and for the first kilometer, it travels with a uniform acceleration. Then, for the next two kilometers, it travels with a uniform speed. Finally, the train decelerates uniformly for another kilometer before coming to rest at station B. If the time for the whole journey is six minutes, draw the v-t graph and determine the maximum speed of the train.

12-47. The particle travels along a straight line with the velocity described by the graph. Construct the a-s graph.

v (m/s)

13+------_=_ 10----I v = s + 7 '--v = 2s + 4 4-

I----+---+--s (m) 3 6

Prob. 12-47

*12-48. The a-s graph for a jeep traveling along a straight road is given for the first 300 m of its motion. Construct the v-s graph. At s = 0, v = O.

-12-49. A particle travels along a curve defined by the equation s = (t3 -3t2 + 2t) m. where t is in seconds. Draw the s -t, v -t, and a -t graphs for the particle for o t 3 s.

12-50. A truck is traveling along the straight line with a velocity described by the graph. Construct the a-s graph for 0 s 1500 ft.

v (fIls) v = 0.6 s3/4 75 �-��--------�

<-------r-----------r-----s(fl) 625 1500

Prob.12-50

12-51. A car starts from rest and travels along a straight road with a velocity described by the graph. Determine the total distance traveled until the car stops. Construct the s-t and a-t graphs.

v(rnls) I .,

30 60 90 Prob.12-51

12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 27

*12-52. A car travels up a hill with the speed shown. • Determine the total distance the car travels until it stops

(t = 60 s). Plot the a-t graph.

v (rnls)

 10 +---------

'-----------+------------"!------1 (s) 30 60

Prob.12-52

-12-53. The snowmobile moves along a straight course according to the v-t graph. Construct the s-t and a-t graphs for the same 50-s time interval. When t = 0, s = O.

v (rnls) (s) 30 50

Prob.12-53

12-54. A motorcyclist at A is traveling at 60 ft/s when he wishes to pass the truck T which is traveling at a constant speed of 60 ft/s. To do so the motorcyclist accelerates at

6 ft/S2 until reaching a maximum speed of 85 ft/s. If he then maintains this speed, determine the time needed for him to reach a point located 100 ft in front of the truck. Draw the v-t and s-t graphs for the motorcycle during this time.

()= 60 fIls

Prob.12-54

28 CHAPTER 12 KINEMATICS OF A PARTICLE

12-5. An airplane traveling at 70 mls lands on a straight runway and has a deceleration described by the graph. Determine the time t' and the distance traveled for it to reach a speed of 5 m/s. Construct the v-t and s-t graphs for this time interval, 0 :0; t :0; t'.

o(m/s2) �

5 t' I I(S) -4

Prob.12-5

*12-56. The position of a cyclist traveling along a straight road is described by the graph. Construct the v-t and a-t graphs.

S (m)

S = -0.625? 27.51 -162.5 137.5

'-"'---+----+-----I (s) 10 20

Prob.12-56

-12-57. The dragster starts from rest and travels along a straight track with an acceleration-deceleration described by the graph. Construct the v-s graph for 0 :0; S :0; s', and determine the distance s' traveled before the dragster again comes to rest.

o(m/s2) � a=o.IS;V s' sCm) 1200

Prob.12-57

12-58. A sports car travels along a straight road with an acceleration-deceleration described by the graph. If the car starts from rest, determine the distance s' the car travels until it stops. Construct the v-s graph for 0 :0; S :0; s'.

6 -j-------,

1000 S' --+-----1f-------+-,--,S(ft) -4

Prob.12-58

12-59. A missile starting from rest travels along a straight track and for 10 s has an acceleration as shown. Draw the v-t graph that describes the motion and find the distance traveled in 10 s.

40 +-------., a = 21 + 20

30--'----�

a = 61

Prob.12-59

*12-60. A motorcyclist starting from rest travels along a straight road and for 10 s has an acceleration as shown. Draw the v-t graph that describes the motion and find the distance traveled in 10 s.

6+------�------�

'"-"""""----+----+---1 (s) 6 10

Prob. 12-60

12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 29

-12-61. The v-t graph of a car while traveling along a road • is shown. Draw the s-t and a-t graphs for the motion.

v (m/s)

20+---r-----------�

Prob.12-61

12-62. The boat travels in a straight line with the acceleration described by the a-s graph. If it starts from rest, construct the v-s graph and determine the boat's maximum speed. What distance S' does it travel before it stops?

6 �.02S+6

150 S' sCm) -4

Prob.12-62

30 CHAPTER 12 KINEMATICS OF A PARTICLE

12-63. The rocket has an acceleration described by the graph. If it starts from rest, construct the v-( and s-( graphs for the motion for the time interval 0 :S ( :S 14s.

'-------i----i----t(s) 9 14

Prob.12-63

*12-64. The jet bike is moving along a straight road with the speed described by the v-s graph. Construct the a-s graph.

v(m/s)

v = 5s1/2

75 - v = -0.2s + 120

15 sCm) 225 525

-12-65. The acceleration of the speed boat starting from rest is described by the graph. Construct the v-s graph.

10 d�� s(ft) 200 500

Prob.12-65

12-6. The boat travels along a straight line with the speed described by the graph. Construct the s-( and a-s graphs. Also, determine the time required for the boat to travel a distance s = 400 m if s = 0 when ( = o.

v(rnls)

80 v = 0.2s

L----,--------------,---s(m) 100 400

(Parte 5 de 6)