Some examples on kinematics of particles.

1. Example 2.1
IF
What are s, v, and a at t = 2 s, 4s
and 6s ?
2.2) What is true about the
kinematics of a particle?
a)The velocity of a particle is always
positive
b)The velocity of a particle is equal
to the slope of the position-time
graph
c) If the position of a particle is
zero, then the velocity must be
zero
d) If the velocity of a particle is
zero, then its acceleration must be
zero
3
2
6 t
t
s 


2. 2.3) Ball tossed with 10 m/s
vertical velocity from
window 20 m above ground.
Determine:
1. velocity and elevation
above ground at time t,
2. highest elevation reached
by ball and corresponding
time, and
3. time when ball will hit the
ground and corresponding
velocity.

3. SOLUTION
Integrate twice to find v(t) and y(t).
Solve for t when velocity equals zero (time for maximum
elevation) and evaluate corresponding altitude.
Solve for t when altitude equals zero (time for ground
impact) and evaluate corresponding velocity.

4. Integrate twice to find v(t) and y(t).
 
  t
v
t
v
dt
dv
a
dt
dv
t
t
v
v
81
.
9
81
.
9
s
m
81
.
9
0
0
2
0










  t
t
v 






 2
s
m
81
.
9
s
m
10
 
    2
2
1
0
0
81
.
9
10
81
.
9
10
81
.
9
10
0
t
t
y
t
y
dt
t
dy
t
v
dt
dy
t
t
y
y










  2
2
s
m
905
.
4
s
m
10
m
20 t
t
t
y 















5. Solve for t when velocity equals zero and evaluate
corresponding altitude.
  0
s
m
81
.
9
s
m
10 2








 t
t
v
s
019
.
1

t
Solve for t when altitude equals zero and evaluate
corresponding velocity.
 
   2
2
2
2
s
019
.
1
s
m
905
.
4
s
019
.
1
s
m
10
m
20
s
m
905
.
4
s
m
10
m
20






























y
t
t
t
y
m
1
.
25

y

6. Solve for t when velocity equals zero and evaluate
corresponding altitude.
• Solve for t when altitude equals zero and evaluate
corresponding velocity.
  0
s
m
905
.
4
s
m
10
m
20 2
2















 t
t
t
y
 
s
28
.
3
necessary
not
s
243
.
1



t
t
 
   
s
28
.
3
s
m
81
.
9
s
m
10
s
28
.
3
s
m
81
.
9
s
m
10
2
2
















v
t
t
v
s
m
2
.
22


v

7. 2.4) Brake mechanism used to reduce gun recoil consists of
piston attached to barrel moving in fixed cylinder filled with
oil. As barrel recoils with initial velocity v0, piston moves and
oil is forced through orifices in piston, causing piston and
cylinder to decelerate at rate proportional to their velocity.
• Determine v(t), x(t), and v(x).
kv
a 

SOLUTION:
• Integrate a = dv/dt = -kv to
find v(t).
• Integrate v(t) = dx/dt to find
x(t).
• Integrate a = v dv/dx = -kv to
find v(x).

8. SOLUTION:
• Integrate a = dv/dt = -kv to find
v(t).
 
0 0
0
ln
v t
v
v t
dv dv
a kv k dt kt
dt v v
      
 
  kt
e
v
t
v 
 0
• Integrate v(t) = dx/dt to find x(t).
 
 
0
0 0
0
0 0
1
kt
t
x t
kt kt
dx
v t v e
dt
dx v e dt x t v e
k

 
 
 
  
 
 
 
   
kt
e
k
v
t
x 

 1
0

9. Integrate a = v dv/dx = -kv to find v(x).
kx
v
v
dx
k
dv
dx
k
dv
kv
dx
dv
v
a
x
v
v












0
0
0
kx
v
v 
 0
• Alternatively,
   










0
0 1
v
t
v
k
v
t
x kx
v
v 
 0
   
0
0 or
v
t
v
e
e
v
t
v kt
kt

 

   
kt
e
k
v
t
x 

 1
0
with
and
then

10. 2.5) A projectile is fired from the edge of a 150-m cliff with an initial
velocity of 180 m/s at an angle of 30°with the horizontal. Neglecting air
resistance, find (a) the horizontal distance from the gun to the point
where the projectile strikes the ground, (b) the greatest elevation above
the ground reached by the projectile.

11. SOLUTION:
Consider the vertical and horizontal motion separately (they are
independent)
Apply equations of motion in y-direction
Apply equations of motion in x-direction
Determine time t for projectile to hit the ground, use this to find the
horizontal distance
Maximum elevation occurs when vy=0

12. Given: (v)o =180 m/s, (y)o =150 m, (a)y = - 9.81 m/s2, (a)x = 0 m/s2
Vertical motion – uniformly accelerated:
Horizontal motion – uniformly accelerated:
Choose positive x to the right as shown

13. Horizontal distance
Projectile strikes the ground at:
Solving for t, we take the positive root
Maximum elevation occurs when vy=0,
Substitute into equation (1) above
Substitute t into equation (4)
Maximum elevation above the ground =

14. 2.6)Pulley D is attached to a collar
which is pulled down at 3 in./s. At
t = 0, collar A starts moving down
from K with constant acceleration
and zero initial velocity. Knowing
that velocity of collar A is 12 in./s
as it passes L, determine the
change in elevation, velocity, and
acceleration of block B when block
A is at L.
SOLUTION:
Define origin at upper
horizontal surface with positive
displacement downward.

15. Collar A has uniformly
accelerated rectilinear
motion. Solve for
acceleration and time t to
reach L.
Pulley D has uniform
rectilinear motion.
Calculate change of position
at time t.
Block B motion is dependent
on motions of collar A and
pulley D. Write motion
relationship and solve for
change of block B position at
time t.
Differentiate motion relation
twice to develop equations for
velocity and acceleration of
block B.

16. SOLUTION:
• Define origin at upper horizontal surface with positive displacement
downward.
• Collar A has uniformly accelerated rectilinear motion. Solve for
acceleration and time t to reach L.
   
 
  2
2
0
2
0
2
s
in.
9
in.
8
2
s
in.
12
2











A
A
A
A
A
A
A
a
a
x
x
a
v
v
 
s
333
.
1
s
in.
9
s
in.
12 2
0




t
t
t
a
v
v A
A
A

17. • Pulley D has uniform rectilinear motion. Calculate change of position
at time t.
 
    in.
4
s
333
.
1
s
in.
3
0
0











D
D
D
D
D
x
x
t
v
x
x
• Block B motion is dependent on motions of
collar A and pulley D. Write motion
relationship and solve for change of block
B position at time t.
Total length of cable remains constant,
     
 
   
   
 
     
  0
in.
4
2
in.
8
0
2
2
2
0
0
0
0
0
0
0















B
B
B
B
D
D
A
A
B
D
A
B
D
A
x
x
x
x
x
x
x
x
x
x
x
x
x
x
  in.
16
0 

 B
B x
x

18. • Differentiate motion relation twice to develop equations for velocity
and acceleration of block B.
0
s
in.
3
2
s
in.
12
0
2
constant
2





















B
B
D
A
B
D
A
v
v
v
v
x
x
x
s
in.
18

B
v
0
s
in.
9
0
2
2











B
B
D
A
a
a
a
a
2
s
in.
9


B
a

19. 2.7) Automobile A is traveling east at the constant speed of 36 km/h. As
automobile A crosses the intersection shown, automobile B starts from
rest 35 m north of the intersection and moves south with a constant
acceleration of 1.2 m/s2. Determine the position, velocity, and
acceleration of B relative to A 5 s after A crosses the intersection.

20. SOLUTION:
Define inertial axes for the system
Determine the position, speed, and acceleration of car A at t = 5 s
Determine the position, speed, and acceleration of car B at t = 5 s
Using vectors equations or a graphical approach, determine the relative
position, velocity, and acceleration

21. Define axes along the road
Given:
vA=36 km/h, aA= 0, (xA)0 = 0
(vB)0= 0, aB= - 1.2 m/s2, (yA)0 = 35 m
Determine motion of Automobile A:
We have uniform motion for A so:
At t = 5 s

22. Determine motion of Automobile B:
We have uniform acceleration for B so:
At t = 5 s

23. We can solve the problems geometrically, and apply the arctangent
relationship:
Physically, a rider in car A would “see” car B travelling south and west.
Or we can solve the problems using vectors to obtain equivalent results:
 
B A B/A
r r r  
B A B/A
v v v  
B A B/A
a a a
20 50
20 50 (m)
 
 
B/A
B/A
j i r
r j i
6 10
6 10 (m/s)
  
  
B/A
B/A
j i v
v j i
2
1.2 0
1.2 (m/s )
  
 
B/A
B/A
j i a
a j

24. 2.8) A motorist is traveling on a curved section of highway of radius
2500 ft at the speed of 60 mi/h. The motorist suddenly applies the
brakes, causing the automobile to slow down at a constant rate.
Knowing that after 8 s the speed has been reduced to 45 mi/h, determine
the acceleration of the automobile immediately after the brakes have
been applied.

25. SOLUTION
Define your coordinate system
Calculate the tangential velocity and tangential acceleration
Calculate the normal acceleration
Determine overall acceleration magnitude after the brakes have been
applied
:

26. :
• Define your coordinate system
et
en
• Determine velocity and acceleration
in the tangential direction
• The deceleration constant, therefore
• Immediately after the brakes are applied,
the speed is still 88 ft/s
2 2 2 2
2.75 3.10
n t
a a a
   

27. :
Quiz
1. A baseball pitching machine “throws” baseballs
with a horizontal velocity v0. If you want the height h
to be 42 in., determine the value of v0.

28. :
Quiz
2. Slider block A moves to the left with a constant velocity of 6 m/s.
Determine the velocity of block B.
• Sketch your system and choose
coordinate system
• Write out constraint equation
• Differentiate the constraint equation
to get velocity

29. Quiz
• Define your coordinate system
• Calculate the tangential velocity
and tangential acceleration
• Determine overall acceleration
magnitude
• Calculate the normal acceleration
3a) The tangential acceleration
of the centrifuge cab is given by
where t is in seconds and at is in
m/s2. If the centrifuge starts from
fest, determine the total
acceleration magnitude of the cab
after 10 seconds.
2
0.5 (m/s )
t
a t


30. Quiz
• Define your coordinate system
• Calculate the angular velocity
after three revolutions
• Calculate the radial and
transverse accelerations
• Determine overall acceleration
magnitude
3b) The angular acceleration of
the centrifuge arm varies
according to
where q is measured in radians. If
the centrifuge starts from rest,
determine the acceleration
magnitude after the gondola has
travelled two full rotations.
2
0.05 (rad/s )
 
