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Zina [86]
3 years ago
10

A car traveling at a velocity v can stop in a minimum distance d. What would be the car's minimum stopping distance if it were t

raveling at a velocity of 2v?
a. 4d
b. 2d
c. 8d
d. √2 d
e. d
Physics
1 answer:
alexira [117]3 years ago
5 0

Answer:

a. 4d.

If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.

Explanation:

Hi there!

The equations of distance and velocity of the car are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x =  position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:

velocity = v0 + a · t      v0 = v

0 = v + a · t

Solving for t:

-v/a = t

Since the acceleration is negative because the car is stopping:

v/a = t

Now replacing t = v/a in the equation of position:

x = x0 + v0 · t + 1/2 · a · t²     (let´s consider x0 = 0)

x = v · (v/a) + 1/2 · (-a) (v/a)²    

x = v²/a - 1/2 · v²/a

x = 1/2 v²/a

At a velocity of v, the stopping distance is 1/2 v²/a = d

Now, let´s do the same calculations with an initial velocity v0 = 2v:

Using the equation of velocity:

velocity = v0 + a · t

0 = 2v - a · t

-2v/-a = t

t = 2v/a

Replacing in the equation of position:

x1 = x0 + v0 · t + 1/2 · a · t²  

x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²

x1 = 4v²/a - 2v²/a

x1 = 2v²/a

x1 = 4(1/2 v²/a)

x1 = 4x

x1 = 4d

If the car travels at a velocity 2v, the minimum stopping distance will be 4d.

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A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are
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Answer:

The magnitude and direction of the acceleration of the particle is a= 0.3296\ \hat{k}\ m/s^2

Explanation:

Given that,

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We need to calculate the acceleration of the particle

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3 years ago
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If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of
kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

<em>F</em> ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

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irga5000 [103]

Answer:

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Reduce everything by dividing them by 5

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For instantaneous velocity, differentiate x(t). That is,

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But dy/dt = velocity.

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V = 0.

and to find instantaneous acceleration, differentiate dv/dt

dv/dt = 60

Therefore, acceleration = 60 m/s^2

4 0
3 years ago
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