Question

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

Answer 1

Explanation:

a1=a2yufjoup]ytu8yyp

Answer 2

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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