# An object is at rest at #(8 ,3 ,8 )# and constantly accelerates at a rate of #7/4 m/s^2# as it moves to point B. If point B is at #(6 ,5 ,2 )#, 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, you can use the kinematic equation:

[ \text{Distance} = \text{Initial velocity} \times \text{Time} + \frac{1}{2} \times \text{Acceleration} \times \text{Time}^2 ]

Given that the initial velocity is zero (object is at rest), the equation simplifies to:

[ \text{Distance} = \frac{1}{2} \times \text{Acceleration} \times \text{Time}^2 ]

Rearrange the equation to solve for time:

[ \text{Time} = \sqrt{\frac{2 \times \text{Distance}}{\text{Acceleration}}} ]

Substitute the coordinates of the points A and B to find the distance:

[ \text{Distance} = \sqrt{(6 - 8)^2 + (5 - 3)^2 + (2 - 8)^2} ]

Then, substitute the values into the time equation:

[ \text{Time} = \sqrt{\frac{2 \times \text{Distance}}{\frac{7}{4}}} ]

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To find the time it takes for the object to reach point B, we can use the kinematic equation:

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

where:

- ( s ) is the displacement,
- ( u ) is the initial velocity (which is assumed to be zero as the object starts from rest),
- ( a ) is the constant acceleration, and
- ( t ) is the time taken.

Given:

- Initial position ( \mathbf{r_i} = (8, 3, 8) )
- Final position ( \mathbf{r_f} = (6, 5, 2) )
- Acceleration ( a = \frac{7}{4} ) m/s²

We can find the displacement ( s ) using the formula:

[ s = |\mathbf{r_f} - \mathbf{r_i}| ]

[ = \sqrt{(6 - 8)^2 + (5 - 3)^2 + (2 - 8)^2} ]

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

[ = \sqrt{4 + 4 + 36} ]

[ = \sqrt{44} ]

Now, we can use the kinematic equation to solve for time:

[ \sqrt{44} = 0 \cdot t + \frac{1}{2} \cdot \frac{7}{4} \cdot t^2 ]

[ \frac{7}{8} t^2 = \sqrt{44} ]

[ t^2 = \frac{8}{7} \cdot \sqrt{44} ]

[ t = \sqrt{\frac{8}{7} \cdot \sqrt{44}} ]

[ t \approx \sqrt{\frac{8}{7} \cdot 6.633} ]

[ t \approx \sqrt{9.408} ]

[ t \approx 3.07 \text{ seconds} ]

Therefore, it will take approximately ( 3.07 ) 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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