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CHAPTER 1
1.1 Using the definition of acceleration:
α = ∆ ∆vt − 60 mph 9. 2 −^ − 0 0 = 88 ft 9 ./2 secsec^ − 0 = 9.57 ft/sec^2
1.2 Differentiate x(t) to obtain the velocity: v(t) = ˙x(t) = − 10 t + 88 ft/sec. Differentiating again yields the acceleration: a(t) = ¨x(t) = −10 ft/sec^2. So v(t) = 0 = − 10 t + 88. Solving this for t yields that: v(t) = 0 at t = 8.8 sec.
1.3 Evaluating x at zero yields x(0) = 5 m.
Differentiating yields ˙x(t) = v(t) = 3t^2 − 2 so that v(0) = −2 m/s. Likewise a(t) = ¨x(t) = 6t so that a(0) = 0. Now at t = 3 sec, x(3) = 27 − 2(3) + 5 = 26 m, v(3) = 27 − 2 = 25 m/s and a(3) = 18 m/s^2. Since the velocity changes sign during this interval, the particle has doubled back and to compute the total distance traveled during the interval you must compute how far it travels before it changes direction and then add this to the distance traveled after the particle has changed direction. The particle changes direction when the velocity is zero, or at the value of t for which v(t) = 3t^2 − 2 = 0, or at time t = 0.8165. The particle first moves from x(0) = 5 m to x(0.8165) = 3.9 m or a distance of 1.1 m.
It then changes direction and moves from 3.9 m to x(3) = 26 m. Thus it travels a total distance of (26 − 3 .9) + 1.1 = 1.1 + 1.1 + 21 = 23.2 m.
1.4 This again is straightforward differentiation: v(t) = ˙x(t) = 2t − 2. This is zero when t satisfies: 2 t − 2 = 0 or t = 1 sec. Next: a(t) = ¨x(t) = ˙v(t) = 2 ft/sec^2 which is constant acceleration. Alternately, the distance traveled can be computed directly from integrating the absolute value of the velocity: d = ∫^03 | 3 t^2 − 2 |dt = 23.178 m
1.5 Solution: v(t) = ˙x(t) = 6 cos 2t m/s. a(t) = ¨x(t) = ˙v(t) = −12 sin(2t) m/sec^2. Setting −12 sin(2t) = 0, yields 2t = π, or t = 0, π/2 s, π, 3 π/ 2 ...nπ/ 2 , for the times for the acceleration to hit zero.
1.6 From the definition, straightforward differentiation yields: xA = (3t^2 + 6t) ft so the vA = (6t + 6) ft/sec and aA = 6 ft/sec^2. Likewise, xB = 3t^3 + 2t so that vB = (9t^2 + 2) ft/sec and aB = 18t ft/sec^2 a)Thus at t = 1 sec: xA(1) = 9 ft, vA(1) = 12 ft/sec, and aA(1) = 6 ft/sec^2 xB (1) = 5 ft, vB (1) = 11 ft/sec, and aB (1) = 18 ft/sec^2 So that A is ahead of B and has the largest velocity, but B is accelerating faster the A. b) Now at t = 2 sec: xA(2) = 24 ft, vA(2) = 18 ft/sec, and aA(2) = 6 ft/sec^2 xB (2) = 28 ft, vB (2) = 38 ft/sec, and aB (2) = 36 ft/sec^2 Now B is ahead of A, and has larger velocity and acceleration
kill and they must be reminded that simple derivations should be something they can do in their head, on tests, while complicated derivatives are more accurately done with software. The plots are also easy to sketch by hand, but if they are inexperienced at plotting using software, it is best to start them off with some simple plots.
a) v(t) = ˙x = 0.6 cos 2t a(t) = ¨x = − 1 .2 sin 2t. b) 3 sin 2t travels a distance of 0.3m in the time 0 to π 4 sec and back another 0.3m returning back to the origin from π 4 to π 2 s. So the total distance traveled is 0.3 + 0.3 = 0.6m (maybe look at the plot first).
c) The position of the mass at t = π 2 s however is x
( (^) π 2
) = 0.3 sin
( (^2) π 2
) = 0.
Note the position at any point is not always the distance traveled, which in b) is shown to be 0.6m.
t 0 ,.01 ..π 2 x t 0.3 .sin 2 .t v t 0.6 .cos 2 .t a t 1.2 .sin 2 .t
0 0.5 1 1.5 2
2
1
1 x t v t a t
t
FIGURE S1.
1.10 Differentiation yields v(t) = 88.92 sin 0. 26 t ft, and a(t) = 23.12 cos 0. 26 t ft/sec^2. The plot of each follows.
t 0 ,0.1 ..12. x t 342. 1 cos 0.26. t v t 88.92 .sin 0.26 .t a t 23.12. cos 0.26 .t
0 2 4 6 8 10 12 14
500
500
1000
x t v t a t
t
FIGURE S1.
1.11 The following code in Matlab computes the velocity from the displacement data and plots it: x=[8 9 11 13 14 15 17 18 22 27 32 37 41 44 46 48 49 49 48 47 46]; t=0;,01:02; n=length (x); v=0x; dt=.01; for i=1:n- v(i+1)=(x(i+1)-x(i)/dt; end v plot(t,v),xlabel(‘tdt or elapsed time’), title (‘velocity versus time’)
Columns 19 through 21 -10000 0 0 EDU>>plot(t,a),xlabel(‘elapsed time’), title (‘acceleration ersus time’)
-1 0 0.05 0.1 0.15 0.2 0.
-0.
0
1
2
3 x 10^4
elapsed time
acceleration versus time
FIGURE S1.11b
1.12 Consider the following Matlab code which uses the central difference to compute the velocity: x=[8 9 11 13 14 15 17 18 22 27 32 37 41 44 46 48 49 49 48 47 46]; t=0:,.01:0.2; n=length (x); v=0x; dt=.01; for i=1:n- v(i)=(x(i+1)-x(i-1))/(2dt); end v plot(t,v),xlabel(’t*dt or elapsed time’),title(velocity versus time’)
This results in the following values: v = Columns 1 through 12 0 150 200 150 0 100 150 150 250 0 450 500 500 450 Columns 13 through 21 350 250 200 150 50 -50 -100 -100 0 And the following plot:
-100 0 0.05 0.1 0.15 0.2 0.
0
100
200
300
400
500
t*dt or elapsed time
velocity versus time
The following is the Mathcad code for solving this problem:
velocity in mm/s versus time in s
i 1
xi 1 xi 1
- ∆t
(^2000) 0.05 0.1 0.15 0.
0
200
400
600
vi
t
i
v
∆ t=0.
1.17 Solution: y(t) = 3t^2 − 20 , so y(0) = −20 m v(t) = ˙y(t) = 6t, so ˙v(0) = 0 m/s a(t) = ¨y(t) = 6, so a(0) = 6 m/s^2
1.18 Solution: x(t) = exp(− 0. 1 · t) · (3 cos t cos(2 · t) + sin(2 · t)), x(0) = 3 v(t) = 1. 7 · exp(−. 1 · t) · cos(2 · t) − 6. 1 · exp(−. 1 · t) · sin(2 · t), v(0) = 1. 7 a(t) = − 12 .36exp(− 0. 1 t) · cos(2t) − 2. 79 · exp(− 0. 1 t) · sin(2t), a(0) = − 12. 37
1.19 Solution: x(t) = 5 · t − exp(−t) · 3 · t, x(0) = 0; v(t) = 5 + 3 · exp(−t) · t − 3 · exp(−t), v(0) = 2; a(t) = − 3 · exp(−t) · t + 6 · exp(−t), a(0) = 6.
1.20 Solution: There is no derivative at t = 5, however the problem may be split using heaviside functions and differentiated over the two intervals.
t 0 ,.01 .. 10 y t 5 .t. Φ 5 t 25 20. sin π .t π .Φ t 5 y1 t 5 .Φ 5 t 20. cos π .t π .Φ t 5 is the velocity y11 t 20 .π^2 .sin π .t π .Φ t 5 is the acceleration
0 5 10
200
200
y11 t y1 t
t
FIGURE S1.
During the 1st^ interval the slope dy/dt is constant. From the plot y(t) = (^204) −− 00 t− = 5t for t = 0 to 5 sec. In the interval t > 5, y(t) = 25 + A sin(ωt + φ) where ω is the frequency and φ is the plane and A is the amplitude of the sine wave. From the plot A = 20, the period T = 2 sec, so that ω = (^2) Tπ = π. To find the phase evaluate y(5) = 25 = 25 + 20(sin(πt + φ) so that πt + φ = nπ, where n is any integer or φ = (n − 5)π, so φ = π will work. Thus:
y(t) =
{ (^5) t 0 < t < 5 20 sin(πt − π) t > 5 y′(t) =
{ (^5 0) < t < 5 20 π cos(πt − π) t > 5 y′′(t) =
{ (^0 0) < t < 5 − 20 π^2 sin(πt − π) t > 5
1.21 Solution: aave = 60 − 5 sec0 mph = 605 mph sec = 125280 ft mile mile hour sec^1 3600 sec1 hour = (12)(5280) 300 ft/sec^2
= 17.6 ft/sec^2 17 .6 ft/sec^2 = 17. 6 3 m. 25 secft 2 = 5.4 m/sec^2 5 .4 m(sec) 9 .81 m/sec^2 = 0.55 g’s or about 55% of a g.
1.22 One answer: a 4 min mile is near a record speed for trained runners so: v = 1 mile 4 min = 14 mile min1 min 60 sec · 5280 ft mile = 22 ft/sec or about 6.71 meter/s, or about 15 mph. On the other hand a sprinter can cover 100 m dash in 10 seconds, or 10 m/s.
1.23 The Matlab solution follows (see problem 1.11 and 1.12 also): %first assign the data to the vector v v = [0 0.2 0.27 0.3 0.3 0.3 0.3 0.3 0.35 0.4 0.5]; n=length(v); %assign the time step t=0:,0.1:11; dt=0.1; a=0.v; x=0.v; % zeros in a and x %using a loop (see statics supplement or student ed. of Matlab)
Next the equivalent Matlab code is given (without the plots as they look the same). This is the same type of code used in problems 1.11, 1.12 and 1.23.
x=[0 0 0 0.02 0.04 0.1 0.18 0.25 0.38 0.7]; t=0:0.1:10; dt=0.1; n=length(x); v=0x; a=0v; for i=1:n- v(i)=(x(i+1)-x(i))/0. a(i)=(v(i+1)-v(i))/0. end plot (t,v), title (‘velocity versus time’) plot (t,a), title (‘acceleration versus time’)
1.25 This is of second order and is linear x(t). The term sin(πt) is nonlinear but in t, not x.
1.26 This is first order and linear in v(t).
1.27 This is first order and linear in v(t)
1.28 This is second order in θ and nonlinear because of the term sin θ(t).
1.29 This is second order in θ(t) and nonlinear because of the terms (dθ/dt)|dθ/dt| and sin θ(t).
1.30 This is first order in v and nonlinear because of the term v^2.
1.31 This is second order and linear in x(t).
1.32 This is still linear of second order in x(t).
1.33 Solution: a(t) = 3t^2 − 4 t, x(0) = 0, v(0) = 0 v(t) = ∫^0 t (3t^2 − 4 t)t = t^3 − 2 t^2 ft/sec x(t) = ∫^0 t (t^3 − 2 t^2 dt = 14 t^4 − 23 t^3 ft
The plot from Mathcad is: t 0 ,0.01.. 2
0 1 2
2
1
0
x t
t
x(t): = 0.25 t^2 - 3 2 _t^3
FIGURE S1.
1.34 Solution: a(t) = 40t cos πt ∫ (^) v 3 dv^ =^
∫ (^) t 0 40 x^ cos^ πxdx v(t) = 3 + (^40) π 2 [cos πt + tπ sin πt − 1] ∫ (^) x 5 dx^ = 3t^ +^ π^402
∫ (^) t 0 (cos^ πx^ +^ xπ^ sin^ πx^ −^ 1)dx x(t) = 5 + 3t + (^40) π 3 sin πt − (^40) π t + (^) π^403 [sin πx − πx cos πx]t 0 x(t) = 5 + (3t − (^40) π t) + (^40) π 3 [2 sin πt − πt cos πx]
b) Equation (1.29) relates constant acceleration to velocity, time and position. When the ball returns to its initial position x = 0 and equation (1.29) becomes 0 = a^02 t 2 + v 0 t + 0. One solution is t = 0 which is the initial state. If t 6 = 0, the relationship becomes t = (^) −v 20 a 0 = 2. 038 ∼ 2 sec
1.37 For constant acceleration v = at + v 0 and v^2 − v^20 = 2ax so
a = v^2 2 −xv 20 = (5×^10 (2)(2)^6 )^2 −(10 4 )^2 = 6. 3 × 1012 m/s^2 , t = v− av 0 = 5 × 610. 36 ×− 101 × 1210 4 = 8. 0 × 10 −^7 s.
1.38 Here a is a function of x, so consider the development of part 2, eq. (1.17) and (1.18) adx = vdv or −kxdx = vdv or −kx 22 |xx 0 = v 22 |x 0 kx^20 = kx^2 or v^2 = kx^20 − kx^2 or v(x) =
√ k(x^20 − x^2 ). The position time relationship can be found from (1.20): t = ∫^0 t dt = ∫^ xx 0 v^ dx(x) = ∫^ xx 0 √k(^ dxx (^20) −x (^2) ) = √^1 k sin−^1
( (^) x x 0
) |xx 0. Rearranging and solving for x(t) yields t =
√ (^1) k sin−^1
( (^) x x 0
) |xx 0 or
kt = sin−^1
( (^) x x 0
) − sin−^1 (1) or
kt + π 2 = sin−^1
( (^) x x 0
) or x(t) = x 0 sin(
kt + π/2) = x 0 cos(
kt).
1.39 As the elevator starts from rest with constant acceleration to its operating speed v^2 = 2ax and v = at or t = v/a = (3 m/s)/25 m/s^2 = 1.2 sec and travels a distance of x = v^2 / 2 a = (3 m/s)^2 /2(2.5 m/s^2 ) = 1.8 m. Which is also the time and distance required to stop the elevator. Hence 2 × 1 .2 s = 2.4 s and 2 × 1 .8 m or 3.6 m are used up in starting up and slowing down, the remaining distance 200 m - 3.6 m or 196.4 m is traveled at a constant velocity of 3 m/s so t = x/v = 196. 4 /3 m/s = 65.5 sec.
The total time is then 1.2 + 05.5 + 1.2 = 67.9 sec or a little over one minute.
1.40 Since a = −c^2 x case 2 applies and
v^2 − v^20 2 =
∫ (^) x x 0 −c
(^2) xdx = −c 2
( x^2 − x^20 2
)
or v^2 = v^20 + c^2 x^20 − c^2 x^2. Substitution of x 0 = 0, x = 10, v 0 = 30, v = 0, yields 0 = 30^2 + 0 − c^2102 or c = 3.
1.41 Given a(x) = −cx^2 x 0 = 0, t 0 = 0, v(0) = v 0 we want to determine v(x). From eq. (1.17) −c ∫^0 x x^2 dx = ∫^ vv 0 vdv = 12 (v^2 − v 02 ) or (^12) (v (^2) − v (^20) ) = − 3 c x (^3) or v(x) = √ v (^20) − 23 c x 3
1.42 Since a is given as a function of velocity, case 3 applies: dx = vdv/f (v) upon integrating x − x 0 = ∫^ vv 0 vdv/(−v) = −v + v 0. Since x 0 = 0 and v = 0 when it comes to rest, x = 750 mm.
1.43 From the problem statement y 0 = 40 km, v 0 = 6000 km/s calculate an expres- sion for y. Here acceleration is a function of position, so equations (1.17)-(1.20) apply. Given a(y) = −g (^0) (RR+^2 y) 2 , g 0 = 9.81 m/s^2 , R = 6370 × 103. At t = 0, y 0 = y(0) = 40 × 103 m, v 0 = v(0) = 6000 m/s Note ymax will occur when v = 0. So compute v. a = dv dt = dv dydy dt = −g (^0) (RR+^2 y) 2. Integrating yields ∫ (^0) v 0 vdv^ =^ −gR^2
∫ (^) ym y 0 (R^ dy+y)^2 =^ g^0 R^2
[ (^1) (R+ym ) −^ (R+^1 y 0 )
] , or (^022) − v 220 = g 0 R 2 [^ R+ (^1) y m −^ R+^1 y 0
] . Thus (^60002 2) = g 0 R 2 [^ (R+ (^1) y 0 )^ −^ (R+^1 y m)
] or
- 552 × 10 −^8 =
[ (^1) R+y 0 −^ R+^1 ym
] . Solving for ym yields ym = 2575.
1.47 Solution: v 0 = 0.6 ft/s, a = −v^3 ft/s^2 Thus this is case 3 on page 19. However a straightforward integration of a = dv/dt yields a(v) = −v^3 = dv dt. Then dt = −dv v 3 and integrating yields t = − (^21) v 2 + 1.389. Rearrange to get v(t) =
( (^1) 2 t+2. 778
) 1 / 2 ft/s. At t = 4 sec, v(4) = 0.305 ft/s.
1.48 Follow example 1, because the acceleration is a function of velocity so case 3 is used. Note that a = dv dt = g − cv^2 or (^) g−dvcv 2 = dt. Integrating both sides using the stated initial conditions yields
t =
∫ (^) v 0
dv g − cv^2 =
c
∫ (^) v v 0
dv g c − v 2 =
c
) 1 2
√ g/c
(^) ln
√ √^ g/c^ +^ v g/c − v
for g c > v^2 and 4(^1 c ) 2 √^1 g/c ln
( (^) v− √g/c v+√g/c
) for v^2 > g/c, from using a table of integrals. Thus there are two possibilities. For g/c > v^2 ; t = 2 √^1 g/c lm
[ √g/c+v √g/c−v
] . Solving for v yields v(t) =
√ (^) g c
( (^) e 2 √gct− 1 e^2 √gct+
) ; g/c > v^2. For g/c < v^2 ,
t = 2 √^1 gc ln v−
√g/c v+^ √g/c and solving for v(t) yields v(t) =
√ (^) g c
( (^) 1+e 2 √gct 1 −e^2 √gct
) ; v^2 > g/c. Note from this second expression the v^2 = g/c results in the expression −e^2
√ 2 ct
e^2
√ 2 ct which has no solution. Thus v cannot reach the value g/c, i.e., v = g/c is the driver’s terminal velocity. Next consider integrating again to calculate x(t), i.e., dx = vdt or (^) ∫ x 0 dx^ =
√ (^) g c
∫ (^) t 0
( e^2 √gct^ − 1 e^2 √gct^ + 1
) dt
=
√ (^) g c
[∫ (^) t 0
e^2 √gct e^2 √gct^ + 1dt^ −
∫ (^) t 0
dt e^2 √gct^ + 1
]
x(t) =^1 c
[ ℓn
( e^2
√gct
) − √gct − 0. 693
]
1.49 This is a free fall problem or uniformly accelerated motion, where the accel- eration is given as g = 32.2 ft/s, and the time traveled can be determined by equation (1.29) with tf as the given. Equation (1.29) becomes x(tf ) = 30 ft = gt
(^2) f 2 +^ v^0 tf^ +^ x^0 Here v 0 = 0, since the ball is dropped, x 0 = 0 taking the window as the starting position and hence gt^2 f 2 = 30 or^ tf^ =
√ 30 / 32 .2 = 1.365 sec, which is the time required to hit the ground. The expression for velocity under uniform or constant acceleration is equation (1.28) or (1.30). From (1.28) v(tf ) = a 0 (tf ) + v 0 or v(1.365) = (32.2)(1.365) = 43.95 ft/sec.
1.50 This is a case of uniform acceleration a 0 = g = 32.2 ft/s^2 , with v 0 up, and tf = 1.71 s. Using eq. (1.29) again with v 0 as the unknown yields v(tf ) = 30 − (g)t
(^2) f 2 + (−v^0 )tf^ + 0 Here −v 0 is used because v 0 is up and we have taken down as positive in writing a plus sign for a 0 (= g). This is consistent with the solution to 1.49. Solving for v 0 yields v 0 =
[ (g 2 )t^2 f − 30
] /tf = 9.987 ft/sec.
1.51 Given a 0 = 0.7g (constant acceleration), vf = 0 (because the car comes to a stop). Convert mph to ft/s (60 mph = 88 ft/s, 45 mph = 66 ft/s, 30 mph = 44 ft/s) and use eq. (1.30) v^2 f = 2a 0 (xf − x 0 ) + v 02 where vf = 0, a 0 = − 0 .7 g (minus because it decelerates), x 0 = 0 (we start our distance measurement t = 0) then xf = (^) −v 220 a 0 = 0 v.4g^02 = (^) (1.4)(32v^20 (ft.2) ft/sec)/^2 sec 2 so that a) xf = 171.78 ft, b) xf = 96.63 ft, c) xf = 42.95 ft.
1.52 Following the solution to 1.52m with a 0 = 0.4g yields
xp = (^) (−2)(−v 020 .4)(32.2) so that a) xf = 300.6 ft, b) xf = 169.1 ft, c) xf = 75.2 ft.
