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baherus [9]
3 years ago
12

Is it possible for the displacement to be greater than the distance traveled? Explain.

Physics
1 answer:
natka813 [3]3 years ago
3 0

Answer:

It is not possible for the magnitude of displacement of an object to be greater than the distance that this object travelled.

Explanation:

The displacement of an object is difference between its final and initial position. On the other hand, the distance that an object travelled would be equal to the length of the path that this object takes when it moved from its initial position to the final position.

The distance an object travelled and the magnitude (length) of its displacement each represent the length of a path between the initial and final position:

  • The distance travelled can represent the length of any path between these two points.
  • On the other hand, the magnitude of displacement can only represent the length of the line segment that directly connects the two points.

On a flat surface, the shortest path between two points is always going to be a straight line segment connecting the two points. If these two points are the initial and final position of the object, the magnitude of displacement would simply represent the length of that straight line segment. It is not possible for the distance travelled to be any shorter than the shortest path between the two points. Therefore, the displacement of an object would always be less than or equal to the distance it travelled.

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Answer:

Initial Velocity = 0 m/s

Final Velocity = 34.6 m/s

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Explanation:

The initial velocity must be zero since, the egg must be at rest initially, before dropping.

<u>Initial Velocity = 0 m/s</u>

Now, for time we use 2nd equation of motion:

h = Vi t + (1/2)gt²

where,

h = Height = 61 m

Vi = Initial Velocity = 0 m/s

g = 9.8 m/s²

t =time = ?

Therefore,

61 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (61 m)(2)/(9.8 m/s²)

t = √(12.45 s²)

<u>t = 3.5 s</u>

Now, for final velocity we will use 1st equation of motion:

Vf = Vi + gt

Vf = 0 m/s + (9.8 m/s²)(3.5 s)

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3 0
3 years ago
A car stops in 120 m. If it has an acceleration of –5m/s 2 , how long did it take to stop
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Answer:

t=240s

Explanation:

Distance=120m

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Let u=x m/s

Using equation v^2-u^2=2as:-

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x=1200m/s

Using now equation v=u+at:-

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8 0
4 years ago
A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 205 Hz. A person on the pl
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Answer:

Explanation:

The question relates to Doppler effect and beat.

The observer is moving towards the reflected sound so apparent frequency will be increased

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v₁ = v₂ = vp as both observer and source have same velocity

f = f₀ x (V + v₁) / (V - v₂)

205 +5 = 205 x (344 +vp)/ ( 344 - vp)

1.0234 = (344 +vp)/ ( 344 - vp)

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12 / 2.0234vp

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