A particle at 9 AM is moving towards the east at 4 ms At 12 noon, it changes its velocity and starts moving towards the north uniformly at 4 ms the average acceleration of the particle during the interval 10 AM to 1 PM is a1, and the average acceleration of the particle during the interval 10 AM to 2 PM is a2 then.
a1=a2
|a1|>|a2|
a1 and a2 are parallel
a1 is inclined to a2
Answers
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Explanation:
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Answer:
the acceleration 1 is greater than the acceleration 2
Explanation:
Acceleration is the rate of change of velocity over time.
Acceleration and velocity are vectors
If the east and north are in the positive direction, the east movement vector decreases to zero, and the north movement vector increases from zero to 4 m / s.
There are 3 hours or 10800 seconds between 10 am and 1 pm
a1 = √ ((-4) ² + 4²) / 10800 = (√32) / 10800m / s² ≈ 4.2x10 ⁻⁴m / s²
10 am to afternoon 14400 seconds by 2 o'clock
Velocity change is still equal to
a2 = √ ((-4) ² + 4²) / 10800 = (√32) / 14400m / s² ≈ 3.9x10 ⁻⁴m / s²
Which indicates that a1>a2.
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