lec01-part a-in class solutions

8/19/2019 Lec01-Part a-In Class Solutions

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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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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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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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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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Solution:

For each distance range the a-s graph can be found

1. Distance range : 0


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Solution:

For each distance range the a-s graph can be found



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sec05.4505050

/50 20005.8

st dsdt ds

vdsdt s ft v s ft

T



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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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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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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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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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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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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

lec01-part a-in class solutions

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