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ratelena [41]
3 years ago
6

A car is traveling at 50 ft/s when the driver notices a stop sign 100 ft ahead and steps on the brake. Assuming that the deceler

ation due to braking is constant, what is the magnitude of the minimum deceleration for which the car would stop right at the stop sign?
Engineering
1 answer:
Naddika [18.5K]3 years ago
4 0

Answer:

minimum deceleration = 12.5 ft/s²

Explanation:

Data provided in the question:

Initial speed of the car, u  = 50 ft/s

Distance to cover, s = 100 ft

Final speed of the car = 0 ft/s             [ ∵ as the car stops ]

Now,

from the Newton's equation of motion

we have  

v² - u² = 2as

a is the acceleration

on substituting the respective values, we get

0² - 50² = 2 × a × 100

or

- 2500 = 200a

or

a = - 12.5 ft/s²

Here, the negative sign depicts the deceleration.

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A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i
Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter d_{o} = 12.8 mm

Final diameter d_{f} = 10.7

Engineering stress  \alpha _{E} = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(d_{o} )²  ; Ag = π/4(d_{f} )²

% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

= ( (d_{f} )² - (d_{o} )²) / (d_{o} )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

E_{E} = 0.431

so we substitute the value of E_{E}  into our initial equation;

True stress  \alpha _{T} = 460 ( 1 +  0.431)

True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

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Answer:

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2. My=3440 ft lb

3. ∑Mz= 0

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