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3 years ago

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The final position of a car can be predicted with the following equation: Position = -2.5 (time) + 25 Where is the car’s positio

n at 3.0 seconds? When would the car reach position 0.0 meters?

1 answer:

nekit [7.7K]3 years ago

8 0

Answer:

The car's position at 3.0 seconds is 17.5 meters. The car will reach 0.0 meters at 10 seconds.

Explanation:

You can rewrite the equation as P= -2.5t+25. t=time and P is equal to position; so you will plug in 3.0 seconds into t. The equation will now look like: P = -2.5(3.0)+25

P = -2.5(3.0) + 25

P = -7.5 + 25

P = 17.5 meters

Now, to find when the car reaches 0.0 meters, we will plug that into P since P = position and will solve for t.

P = -2.5t + 25

0.0 = -2.5t + 25

-25 -25

-25 = -2.5t

-25/-2.5 = -2.5t/-2.5

10 = t

So the car will reach 0.0 meters at 10 seconds.

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A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy a

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Answer:

Explanation:

From the given question:

Using the distortion energy theory to determine the factors of safety FOS can be expressed by the relation:

$\dfrac{Syt}{FOS}= \sqrt{ \sigma x^2+\sigma y^2-\sigma x \sigma y+3 \tau_{xy^2}}$

where; syt = strength in tension and compression = 350 MPa

The maximum shear stress theory can be expressed as:

$\tau_{max} = \dfrac{Syt}{2FOS}$

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$\tau_{max} =\sqrt{ (\dfrac{\sigma x-\sigma y}{2})^2+ \tau _{xy^2$

a. Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{94^2+0^2-94*0+3 (-75)^2}}$

$\dfrac{350}{FOS}=160.35$

${FOS}=\dfrac{350}{160.35}$

FOS = 2.183

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{94-0}{2})^2+ (-75)^2$

$\dfrac{350}{2 FOS} =88.51$

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${ FOS} =\dfrac{350}{2 \times 88.51}$

FOS = 1.977

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Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 110^2+100^2-110*100+3(0)^2}$

$\dfrac{350}{FOS}= \sqrt{ 12100+10000-11000$

$\dfrac{350}{FOS}=105.3565$

$FOS=\dfrac{350}{105.3565}$

FOS =3.322

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{110-100}{2})^2+ (0)^2$

$\dfrac{350}{2 FOS} ={ (\dfrac{110-100}{2})^2$

$\dfrac{350}{2 FOS} =25$

FOS = 350/2×25

FOS = 350/50

FOS = 70

c. σx = 90 MPa, σy = 20 MPa, τxy =−20 MPa

Using distortion- energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 90^2+20^2-90*20+3(-20)^2}$

$\dfrac{350}{FOS}= \sqrt{ 8100+400-1800+1200}$

$\dfrac{350}{FOS}= 88.88$

FOS = 350/88.88

FOS = 3.939

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{90-20}{2})^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ (35)^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ 1225+ 400$

$\dfrac{350}{2 FOS} =40.31$

$FOS} =\dfrac{350}{2*40.31}$

FOS = 4.341

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