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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.
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-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?
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.