Solution Manual For Vector Mechanics For PDF
Solution Manual For Vector Mechanics For PDF
Solution Manual For Vector Mechanics For PDF
SOLUTION
Position: x 0.5t 3 t 2 2t
dx
Velocity: v 1.5t 2 2t 2
dt
dv
Acceleration: a 3t 2
dt
x 0.5 5 52 2 5
3
At t 5 s, x 97.5 ft
v 1.5 5 2 5 2
2
v 49.5 ft/s
a 3 5 2 a 17 ft/s 2
The motion of a particle is defined by the relation x 2t 3 9t 2 12t 10, 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
x 2t 3 9t 2 12t 10
dx
Differentiating, v 6t 2 18t 12 6(t 2 3t 2)
dt
6(t 2)(t 1)
dv
a 12t 18
dt
So v 0 at t 1 s and t 2 s.
At t 1 s, x1 2 9 12 10 15 t 1.000 s
a1 12 18 6 x1 15.00 ft
a1 6.00 ft/s2
At t 2 s,
x2 14.00 ft
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
SOLUTION
x 60e4.8t sin16t
dx
v 60(4.8)e 4.8t sin16t 60(16)e 4.8t cos16t
dt
v 288e 4.8t sin16t 960e 4.8t cos16t
dv
a 1382.4e4.8t sin16t 4608e4.8t cos16t
dt
4608e4.8t cos16t 15360e4.8t sin16t
a 13977.6e4.8t sin16t 9216e4.8 cos16t
(a) At t 0, x0 0 x0 0 mm
v0.3 (288)(0.23692)(0.99616)
(960)(0.23692)(0.08750) 87.9 v0.3 87.9 mm/s
a0.3 (13977.6)(0.23692)(0.99616)
(9216)(0.23692)(0.08750) 3108 a0.3 3110 mm/s 2
or 3.11 m/s 2
The motion of a particle is defined by the relation x 6t 4 2t 3 12t 2 3t 3, where x and t are expressed in
meters and seconds, respectively. Determine the time, the position, and the velocity when a 0.
SOLUTION
We have x 6t 4 2t 3 12t 2 3t 3
dx
Then v 24t 3 6t 2 24t 3
dt
dv
and a 72t 2 12t 24
dt
2 1
or t s and t s (Reject) t 0.667 s
3 2
4 3 2
2 2 2 2 2
At t s: x2/3 6 2 12 3 3 or x2/3 0.259 m
3 3 3 3 3
3 2
2 2 2
v2/3 24 6 24 3 or v2/3 8.56 m/s
3 3 3
The motion of a particle is defined by the relation x t 3 9t 2 24t 8, where x and t are expressed in inches
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 t 3 9t 2 24t 8
dx
Then v 3t 2 18t 24
dt
dv
and a 6 t 18
dt
2 s t 3 s: v0
Now
At t 0: x0 8 in.
SOLUTION
Position: x t 0.5t 3 3t 2 3t 2
dx
Velocity: v t
dt
v t 1.5t 2 6t 3
x 2 0.5 2 3 2 3 2 2
3 2
(b) Position at t=2 s
x 2 0 m
To find total distance note that car changes direction at t=0.586 s
x 0 0.5 0 3 0 3 0 2
3 2
Position at t=0 s
x 0 2
x 0.586 0.5 0.586 3 0.586 3 0.586 2
3 2
Position at t=0.586 s
x 0.586 2.828 m
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
Total distance traveled 0.828 m 2.828 m Total distance 3.656 m
The motion of a particle is defined by the relation x t t 2 , where x and t are expressed in feet and
2 3
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
x t t 2 t 2
3
Position:
dx
Velocity: v t
dt
v t 2t 3 t 2
2
v t 2t 3 t 2 4t 4
3t 2 14t 12
x 0 0 0 2
2 3
Position at t=0 s
x 0 8 ft
x 4 4 4 2
2 3
Position at t=4 s
x 4 8 ft
(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.
Total distance traveled 6.065 ft 6.944 ft 0.879 ft Total distance 13.888 ft.
SOLUTION
a 10 ft/s 2
(a) Velocity at x 0.
dv
v a 10
dx
0 xf
v0
vdv 0
(10)dx
v02
0 10 x f (10)(300)
2
v02 6000 v0 77.5 ft/s
(b) Time to stop.
dv
a 10
dx
0 tf
v0
dv 0
10dt
0 v0 10t f
v0 77.5
tf t f 7.75 s
10 10
The acceleration of a particle is defined by the relation a 3e 0.2 t , where a and t are expressed in ft/s 2 and
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
dv
Velocity: a dv adt
dt
v t
dv adt
v0 0
t
t
v vo
0
3e 0.2t dt v 0= 15 e0.2t
0
v 15 1 e0.2t ft/s
dx
Position: v dx vdt
dt
x t
x0
dx vdt
0
t
t
x xo 15 1 e 0.2t dt x 0=15 t 5 e 0.2t
0
0
x 15 t 5 e 0.2t 75 ft
Velocity at t=0.5 s v 15 1 e 0.20.5
ft/s
v 0.5 1.427 ft/s
Position at t=0.5 s
x 15 0.5 5 e 0.20.5 75 ft
x 0.5 0.363 ft.
The acceleration of a particle is directly proportional to the square of the time t. When t 0, the particle is
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 kt 2 k constant
dv
Now a kt 2
dt
v t
At t 6 s, v 18 m/s: 18
dv 6
kt 2 dt
1
or v 18 k (t 3 216)
3
1
or v 18 k (t 3 216)(m/s)
3
dx 1
Also v 18 k (t 3 216)
dt 3
x t 1
At t 0, x 24 m: 24
dx 0
18 k (t 3 216) dt
3
1 1
or x 24 18t k t 4 216t
3 4
Now
1 1
At t 6 s, x 96 m: 96 24 18(6) k (6) 4 216(6)
3 4
1
or k m/s 4
9
1 1 1
Then x 24 18t t 4 216t
3 9 4
1 4
or x(t ) t 10t 24
108
11
and v 18 (t 3 216)
3 9
1 3
or v(t ) t 10
27
The acceleration of a particle is defined by the relation a kt 2 . (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 0 when t 2 s.
SOLUTION
a kt 2 (1)
dv
a kt 2
dt
t 0, v 8 m/s and t 2 s, v 8 ft/s
8 2
(a) 8
dv 0
kt 2 dt
1
8 (8) k (2)3 k 6.00 m/s 4
3
dv
a 6t 2 a 6t 2
dt
v t
t 0, v 8 m/s: 8
dv 0
6t 2 dt
1
v (8) 6(t )3 v 2t 3 8
3
dx
v 2t 3 8
dt
t
x t 1 4
t 2 s, x 0: 0
dx 2
(2t 3 8) dt; x
2
t 8t
2
1 1
x t 4 8t (2)4 8(2)
2 2
1 1 4
x t 4 8t 8 16 x t 8t 8
2 2
SOLUTION
t t 1.8 t
v v0 0 a dt 1.8 0 sin kt dt k cos kt 0
1.8
v 0.6 (cos kt 1) 0.6 cos kt 0.6
3
v 0.6cos kt m/s
x t
dx
Position: v
dt
dx vdt
x0
dx vdt
0
t t 0.6 t
x x0 0 v dt 0.6 0 cos kt dt sin kt
k 0
0.6
x0 (sin kt 0) 0.2sin kt
3
x 0.2sin kt m
When t =0.5 s, kt (3)(0.5) 1.5 rad
For the scotch yoke mechanism shown, the acceleration of Point A is defined by the
relation a 1.08sin kt 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
t t
Integrate: v v0 1.08 0 sin kt dt 1.44 0 cos kt dt
1.08 t 1.44 t
v 0.36 cos kt sin kt
k 0 k 0
1.08 1.44
(cos 3t 1) (sin 3t 0)
3 3
0.36 cos 3t 0.36 0.48sin 3t
t t t
x x0 0 v dt 0.36 0 cos kt dt 0.48 0 sin kt dt
0.36 t 0.48 t
x 0.16 sin kt cos kt
k 0 k 0
0.36 0.48
(sin 3t 0) (cos 3t 1)
3 3
0.12sin 3t 0.16 cos 3t 0.16
x 0.12sin 3t 0.16 cos 3t m
Evaluate at t 0.5 s x 0.12sin1.5 0.16 cos1.5 0.1310 m x 131.0 mm
SOLUTION
vdv
a k x
dx
Separate and integrate.
vf xf
v0
vdv 0
k x dx
xf
1 2 1 2 1 1
v f v0 kx 2 k x 2f
2 2 2 2
0
1 1
0 (4)2 k (0.02)2 k 40, 000 s 2
2 2
Maximum acceleration.
a 800 m/s 2
SOLUTION
First note
4 4
When x ft, v 0: 0 (900 ft/s) k ft
12 12
1
or k 2700
s
(a) We have v v0 kx
dv d
Then a (v0 kx) kv
dt dt
or a k (v0 kx)
1
At t 0: a 2700 (900 ft/s 0)
s
dx
(b) We have v v0 kx
dt
x dx t
At t 0, x 0: 0 v0 kx
dt
0
1
or [ln(v0 kx)]0x t
k
1 v0 1 1
or t ln ln
k v0 kx k 1 vk x
0
1 1
When x 3.9 in.: t ln
2700 1s 1 900 ft/s 12 ft
2700 1/s 3.9
or t 1.366 103 s
The acceleration of a particle is defined by the relation a k/x. It has been experimentally determined that
v 15 ft/s 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.
v x dx
15
vdv k
0.6 x
v x
1 2
v k ln x
2 15 0.6
1 2 1 x
v (15) 2 k ln (1)
2 2 0.6
When v 9 ft/s, x 1.2 ft
1 2 1 1.2
(9) (15) 2 k ln
2 2 0.6
Solve for k.
k 103.874 ft 2 /s 2
(a) Velocity when x 65 ft.
1 2 1 1.5
v (15) 2 103.874 ln
2 2 0.6
v 5.89 ft/s
(b) Position when for v 0,
1 x
0 (15) 2 103.874 ln
2 0.6
x
ln 1.083
0.6
x 1.772 ft
SOLUTION
k
The maximum velocity occurs when a 0. 0 9.81
xm2
k 4 104
xm2 40.775 106 m 2 xm 0.0063855 m
9.81 9.81
The acceleration is given as a function of x.
dv k
v a 9.81 2
dx x
Separate variables and integrate:
k dx
vdv 9.81dx
x2
v x x dx
0
vdv 9.81 x0
dx k x0 x2
1 2 1 1
v 9.81( x x0 ) k
2 x x0
1 2 1 1
vm 9.81( xm x0 ) k
2 xm x0
1 1
9.81(0.0063855 0.004) (4 104 )
0.0063855 0.004
0.023402 0.037358 0.013956 m 2 /s 2
4 104
am 9.81 am 15.19 m/s 2
(0.004)2
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 0, determine (a) the velocity of the particle when x 1 m, (b) the position where the velocity is
maximum, (c) the maximum velocity.
SOLUTION
dv x
We have v a 0.1 sin
dx 0.8
v x x
When x 0, v 1 m/s: 1
vdv 0.1 sin
0
dx
0.8
x
1 2 x
or (v 1) 0.1x 0.8 cos
2 0.8 0
1 2 x
or v 0.1x 0.8 cos 0.3
2 0.8
1 2 1
(a) When x 1 m: v 0.1(1) 0.8 cos 0.3
2 0.8
or v 0.323 m/s
x
(b) When v vmax, a 0: 0.1 sin 0
0.8
or x 0.080 134 m x 0.0801 m
SOLUTION
dv lx
Since a is function of x, av 100 x
dx l 2 x2
Separate variables and integrate:
vf 0 lx
v0
vdv 100 x
x0
l x2
2
dx
0
1 2 1 2 x2
v f v0 100 l l 2 x2
2 2 2 x0
1 2 x2
v f 0 100 0 l 2 l l 2 x02
2 2
1 2 100 2
vf (l x02 l 2 2l l 2 x02 )
2 2
100
( l 2 x02 l ) 2
2
v f 10( l 2 x02 l )
The acceleration of a particle is defined by the relation a k 1 e x , where k is a constant. Knowing that
the velocity of the particle is v 9 m/s when x 3 m and that the particle comes to rest at the origin,
determine (a) the value of k, (b) the velocity of the particle when x 2 m.
SOLUTION
Acceleration:
a k 1 e x
Given : at x 3 m, v 9 m/s
at x 0 m, v 0 m/s
adx vdv
k 1 e x
dx vdv
Integrate using x=-3 m and v=9 m/s as the lower limits of the integrals
x v
k 1 e dx vdv
x
3 9
x v
k x e x
1 2
v
2 9
3
Velocity:
k x e x 3 e3 12 v 2 1
(9) 2
2
(1)
(a) Now substitute v=0 m/s and x=0 m into (1) and solve for k
k 0 e 0 3 e3 12 0 2 1
(9)2
2
k 2.52 m 2 /s 2
(b) Find velocity when x= -2 m using the equation (1) and the value of k
2.518 2 e 2 3 e3 12 v 2 1
(9) 2
2
v 4.70 m/s
SOLUTION
vdv
a 0.1(v 2 16)1/2 (1)
dx
Separate and integrate.
v vdv x
0
v 2 16
0
0.1 dx
v
(v 2 16)1/2 0.1 x
0
2 1/2
(v 16) 4 0.1 x
x 10[(v 2 16)1/2 4] (2)
(a) v 3 ft/s.
v 2 16 19.36
v 2 3.36ft 2 /s2 v 1.833 ft/s
SOLUTION
dv
a 10 0.8v
dt
Separate and integrate:
v dv t
v0 10 0.8v
0
dt
v
1
ln(10 0.8v) t
0.8 v0
10 0.8v
ln 0.8t
10 0.8v0
(a) At t 35 s,
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
dv
(a) We have v a k v
dx
so that v dv k dx
v x
When x 0, v 81 m/s: 81
v dv 0
k dx
2 3/2 v
or [v ]81 kx
3
2 3/2
or [v 729] kx
3
2
When x 18 m, v 36 m/s: (363/2 729) k (18)
3
or k 19 m/s 2
Finally
2 3/2
When x 20 m: (v 729) 19(20)
3
or v
2[ v ]81 19t
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
x 0 with a velocity of 16 mm/s, and when x 6 mm the velocity is observed to be 4 mm/s. Determine
(a) the velocity of the particle when x 5 mm, (b) the time at which the velocity of the particle is 9 mm/s.
SOLUTION
adx vdv
2.5
kv dx vdv
Integrate using x=-0 m and v=16 mm/s as the lower limits of the integrals
x v
0
-kdx v 3 2 dv
16
x
v
kx 2v 1 2
16
0
1
Velocity and position: kx 2v 1 2 (1)
2
Now substitute v=4 mm/s and x=6 mm into (1) and solve for k
1
k 6 41 2
2
k 0.0833 mm 3 2 s1 2
(a) Find velocity when x= 5 mm using the equation (1) and the value of k
1
k 5 2v 1 2
2
v 4.76 mm/s
dv
(b) a or adt dv
dt
dv
a or adt dv
dt
Separate variables kdt v 2.5 dv
Integrate using t=0 and v=16 mm/s as the lower limits of the integrals
t v
-kdt v
5 2
dv
0 16
t v
2
kt v 3 2
0 3 16
2 3 2 1
Velocity and time: kt v (2)
3 96
Find time when v= 9 mm/s using the equation (2) and the value of k
2 3 2 1
kt 9
3 96
t 0.171 s
SOLUTION
vdv
Separate variables dx
A Cv 2
Integrate starting from rest and traveling a distance xf with a final velocity vf.
xf vf
vdv
dx A Cv
0 0
2
vf
x
x 0f
1
2C
ln A Cv 2 0
xf
1
2C
ln A v 2f 21C ln A
1 A
xf ln
2C A Cv 2f
2 Cx f
Cv 2f e
Next solve for A A 2 Cx f
1 e