HW 2 Solutions

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5

Distance Traveled: Time for car A to achives can be obtained byapplying Eq. 12–4.

The distance car A travels for this part of motion can be determined by applying Eq. 12–6.

For the second part of motion, car A travels with a constant velocity of and the distance traveled in ( is the total time) is

Car B travels in the opposite direction with a constant velocity of andthe distance traveled in is

It is required that

The distance traveled by car A is

Ans.sA = s1 + s2 = 533.33 + 80(46.67 - 13.33) = 3200 ft

t1 = 46.67 s

533.33 + 80(t1 - 13.33) + 60t1 = 6000

s1 + s2 + s3 = 6000

A :+ B s3 = yt1 = 60t1

t1

y = 60 ft>s

A :+ B s2 = yt¿ = 80(t1 - 13.33)

t1t¿ = (t1 - 13.33) sy = 80 ft>s

s1 = 533.33 ft

802= 0 + 2(6)(s1 - 0)

A :+ B y2= y0

2+ 2ac (s - s0)

t = 13.33 s

80 = 0 + 6t

A :+ B y = y0 + ac t

y = 80 ft>s

12–10. Car A starts from rest at and travels along astraight road with a constant acceleration of until itreaches a speed of . Afterwards it maintains thisspeed. Also, when , car B located 6000 ft down theroad is traveling towards A at a constant speed of .Determine the distance traveled by car A when they passeach other.

60 ft>st = 0

80 ft>s6 ft>s2

t = 0

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A B

6000 ft

60 ft/s

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13


Velocity: The velocity of the particle can be related to its position by applying Eq. 12–3.

[1]

Position: The position of the particle can be related to the time by applying Eq. 12–1.

When ,

Choose the root greater than Ans.

Substitute into Eq. [1] yields

Ans.y =

816(11.94) - 159

= 0.250 m>s

s = 11.94 m

10 m s = 11.94 m = 11.9 m

8s2- 159s + 758 = 0

8(4) = 8s2- 159s + 790

t = 4 s

8t = 8s2- 159s + 790

L

t

0dt =

L

s

10m 18

(16s - 159) ds

dt =

dsy

y =

816s - 159

s - 10 =

12y

-

116

L

s

10mds =

L

y

8m>s-

dy

2y2

ds =

ydya

12–22. A particle moving along a straight line is subjectedto a deceleration , where is in . If ithas a velocity and a position when

, determine its velocity and position when .t = 4 st = 0s = 10 mv = 8 m>s

m>sva = (-2v3) m>s2

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62


Velocity: The x and y components of the box’s velocity can be related by taking thefirst time derivative of the path’s equation using the chain rule.

or

At , . Thus,

Ans.

Acceleration: The x and y components of the box’s acceleration can be obtained bytaking the second time derivative of the path’s equation using the chain rule.

or

At , and . Thus,

Ans.ay = 0.1 C(-3)2+ 5(-1.5) D = 0.15 m>s2 c

ax = -1.5 m>s2vx = -3 m>sx = 5 m

ay = 0.1 Avx 2

+ xax B

y = 0.1[x#

x#

+ xx] = 0.1 Ax# 2+ xx B

vy = 0.1(5)(-3) = -1.5 m>s = 1.5 m>s T

vx = -3 m>sx = 5 m

vy = 0.1xvx

y#

= 0.1xx#

y = 0.05x2

*12–76. The box slides down the slope described by theequation m, where is in meters. If the box hasx components of velocity and acceleration of and at , determine the y componentsof the velocity and the acceleration of the box at this instant.

x = 5 max = –1.5 m>s2vx = –3 m>s

xy = (0.05x2)y

x

y � 0.05 x2

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72


x-Motion: For the motion of the first projectile, , ,and . Thus,

(1)

For the motion of the second projectile, , , and .Thus,

(2)

y-Motion: For the motion of the first projectile, , ,and . Thus,

(3)

For the motion of the second projectile, , , and. Thus,

(4)

Equating Eqs. (1) and (2),

(5)

Equating Eqs. (3) and (4),

(6)t1 =

30 sin u + 1.2262560 sin u - 47.06

(60 sin u - 47.06)t1 = 30 sin u + 1.22625

51.96t1 - 4.905t1 2

= (60 sin u)t1 - 30 sin u - 4.905t1 2

+ 4.905t1 - 1.22625

t1 =

cos u2 cos u - 1

30t1 = 60 cos u(t1 - 0.5)

y = (60 sin u)t1 - 30 sin u - 4.905 t1 2

+ 4.905t1 - 1.22625

y = 0 + 60 sin u(t1 - 0.5) +

12

(-9.81)(t1 - 0.5)2

A + c B y = y0 + vyt +

12

ayt2

ay = -g = -9.81 m>s2y0 = 0vy = 60 sin u

y = 51.96t1 - 4.905t1 2

y = 0 + 51.96t1 +

12

(-9.81)t1 2

A + c B y = y0 + vyt +

12

ayt2

ay = -g = -9.81 m>s2y0 = 0vy = 60 sin 60° = 51.96 m>s

x = 0 + 60 cos u(t1 - 0.5)

A :+ B x = x0 + vxt

t = t1 - 0.5x0 = 0vx = 60 cos u

x = 0 + 30t1

A :+ B x = x0 + vxt

t = t1

x0 = 0vx = 60 cos 60° = 30 m>s

12–90. A projectile is fired with a speed of atan angle of . A second projectile is then fired with thesame speed 0.5 s later. Determine the angle of the secondprojectile so that the two projectiles collide. At whatposition (x, y) will this happen?

u

60°v = 60 m>s

y

y

xx

v � 60 m/s

v � 60 m/s60�u

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73

12–90. Continued


Equating Eqs. (5) and (6) yields

Solving by trial and error,

Ans.

Substituting this result into Eq. (5) (or Eq. (6)),

Substituting this result into Eqs. (1) and (3),

Ans.

Ans.y = 51.96(7.3998) - 4.905 A7.39982 B = 116 m

x = 30(7.3998) = 222 m

t1 =

cos 57.57°2 cos 57.57° - 1

= 7.3998 s

u = 57.57° = 57.6°

49.51 cos u - 30 sin u = 1.22625

cos u2 cos u - 1

=

30 sin u + 1.2262560 sin u - 47.06

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78


Vertical Motion: The vertical component of initial velocity for the football is. The initial and final vertical positions are

and , respectively.

Horizontal Motion: The horizontal component of velocity for the baseball is. The initial and final horizontal positions are

and , respectively.

The distance for which player B must travel in order to catch the baseball is

Ans.

Player B is required to run at a same speed as the horizontal component of velocityof the baseball in order to catch it.

Ans.yB = 40 cos 60° = 20.0 ft>s

d = R - 15 = 43.03 - 15 = 28.0 ft

R = 0 + 20.0(2.152) = 43.03 ft

A :+ B sx = (s0)x + (y0)x t

sx = R(s0)x = 0(y0)x = 40 cos 60° = 20.0 ft>s

t = 2.152 s

0 = 0 + 34.64t +

12

(-32.2)t2

(+ c) sy = (s0)y + (y0)y t +

12

(ac)y t2

sy = 0(s0)y = 0(y0)y = 40 sin 60° = 34.64 ft>s

*12–96. The baseball player A hits the baseball withand . When the ball is directly above

of player B he begins to run under it. Determine theconstant speed and the distance d at which B must run inorder to make the catch at the same elevation at which theball was hit.

vB

uA = 60°vA = 40 ft>s vA � 40 ft/s

AvB

Au B C

15 ft d

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HW 2 Solutions