# An object is at rest at #(8 ,5 ,4 )# and constantly accelerates at a rate of #1/4 m/s^2# as it moves to point B. If point B is at #(2 ,4 ,5 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

The time is

Apply the equation of motion

Therefore,

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To find the time it takes for the object to reach point B, we can use the formula for displacement under constant acceleration:

[ s = ut + \frac{1}{2}at^2 ]

Where:

- ( s ) is the displacement
- ( u ) is the initial velocity (since the object is at rest, ( u = 0 ))
- ( a ) is the acceleration
- ( t ) is the time taken

Given that the initial velocity ( u = 0 ) and the acceleration ( a = \frac{1}{4} , \text{m/s}^2 ), and the displacement from point A to point B is given by the differences in their respective coordinates:

[ s = \sqrt{(2 - 8)^2 + (4 - 5)^2 + (5 - 4)^2} ]

[ s = \sqrt{(-6)^2 + (-1)^2 + (1)^2} ]

[ s = \sqrt{36 + 1 + 1} ]

[ s = \sqrt{38} ]

[ s \approx 6.164 , \text{m} ]

Substituting the values into the displacement equation:

[ 6.164 = 0 \times t + \frac{1}{2} \times \frac{1}{4} \times t^2 ]

[ 6.164 = \frac{1}{8}t^2 ]

[ t^2 = 6.164 \times 8 ]

[ t^2 = 49.312 ]

[ t = \sqrt{49.312} ]

[ t \approx 7.02 , \text{s} ]

So, it will take approximately 7.02 seconds for the object to reach point B.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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