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LEC01-Kinematics of Particles Solutions: Example 1
Tests reveal that a normal driver takes about 0.75s before he or she can
react to a situation to avoid a collision. It takes about 3s for a driver having
0.1% alcohol in his system to do the same. If such drivers are traveling on a
straight road at 30 mph (44 ft/s) and their cars can decelerate at 2ft/s2 ,
determine the shortest stopping distance d for each from the moment they
see the pedestrians.
Moral: If you must drink, please don’t drive!
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LEC01-Kinematics of Particles Solutions: Example 1
s0(reaction)
d (braking)
Plan:
Use rectilinear Kinematics of particle
= , = → =
=
→ = →
=
Solution:
Normal driver:
1. Reaction travel distance (s0)
ft s s ft vt s 3375.0/440 2. Braking distance d(constant deceleration) final velocity will be zero
0
2
0
2tan
200
s savvadsvdv s
s
t consa
v
v
ft d ft d s ft s ft 51733/22/440 2
22
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LEC01-Kinematics of Particles Solutions: Example 1
s0(reaction)
d (braking)
Solution:
Drunk driver:
1. Reaction travel distance (s0)
ft s s ft vt s 1323/440
2. Braking distance d(constant deceleration) final velocity will be zero
ft d ft d s ft s ft 616132/22/440 2
22
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LEC01-Kinematics of Particles Solutions: Example 2
The v –s graph describing the motion of a motorcycle is shown. Construct
the a –s graph of the motion and determine the time needed for the
motorcycle to reach the position s=400ft
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LEC01-Kinematics of Particles Solutions: Example 2
Solution:
For each distance range the a-s graph can be found
1. Distance range : 0
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LEC01-Kinematics of Particles Solutions: Example 2
Solution:
For each distance range the a-s graph can be found
1. Distance range : 0
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LEC01-Kinematics of Particles Solutions: Example 2
sec05.4505050
/50 20005.8
st dsdt ds
vdsdt s ft v s ft
T
2. Distance range : 200
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LEC01-Kinematics of Particles Solutions: Example 3
In a handling test, a car is driven through the slalom course shown. It is
assumed that the car path is sinusoidal and that the maximum lateral
acceleration is 0.7g. If the testers wish to design a slalom through which the
maximum speed is 80 km/h, what cone spacing L should be used?
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LEC01-Kinematics of Particles Solutions: Example 3
Solution:
1. The car sinusoidal motion can be
defined as
Plan:
The maximum of the normal acceleration is provided, using normal-tangent
kinematic to find a curvature that cause this acceleration at the maxvelocity. Based on the obtained curvature, the cone spacing can be arranged
2. Lateral (normal) acceleration:
2va
n
T=2L
LA
x
y
x A y sin
LT
2
an
at
o
To reach max lateral acceleration the radius
of the curvature must be minimized
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LEC01-Kinematics of Particles Solutions: Example 3
3. The minimum of the curvature radius locates at T=π/4 and this radius can
be found from
at T=π /4
Maximum lateral acceleration is 0.7g
x Adx yd
x Adxdy
x A
x A
xd yd
dxdy
sin/
cos/
sin
cos1
/
/1
222
2
2/3222
22
2/32
A
T
AT
T A
T
T A
2
2
2
2
2/3
222
min
4
1
4
2sin
4
2cos1
mT
LmT
mT
kmhv sm g a
n
1.4623.92
34/
6.3/80/81.97.07.0
22
2
min
2
2
max
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LEC01-Kinematics of Particles Solutions: Example 4
A tracking radar lies in the vertical plane of the
path of a rocket which is coasting in unpowered
flight above the atmosphere. For the instant
when θ = 30°, the tracking data give r = 25(104)
ft, ṙ = 4000 ft/sec, and ̇ = 0.80 deg/sec. Theacceleration of the rocket is due only to
gravitational attraction and for its particular
altitude is 31.4 ft/sec2 vertically down. For these
conditions determine the velocity v of the rocket
and the values of r ̈and .
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LEC01-Kinematics of Particles Solutions: Example 4
Solution:
1. The component of velocity in cylindrical sys
Plan:
• Unpowered (no thrust) and above
atmosphere (no aerodynamic drag)• Gravity is the only source of kinematic
changes
• Use cylindrical (polar) coordinate system
s ft s ft ft vvv
s ft r v
s ft vr v
r
r r
/5310/3490/4000
/3490180
8.01025
/4000
2222
4
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LEC01-Kinematics of Particles Solutions: Example 4
Solution:
2. The rocket acceleration is due to gravity
2
2
/7.1530sin4.31
/2.2730cos4.31
s ft a
s ft ar
Acceleration in polar coordinate sys
24
24
22
2
42
/1084.3
/7.15/180
8.04000210252
/5.21/2.27/180
8.01025
srad
s ft srad mmr r a
s ft r s ft srad mr r r ar