# An object is at rest at #(7 ,3 ,9 )# and constantly accelerates at a rate of #7/4 m/s^2# as it moves to point B. If point B is at #(2 ,5 ,0 )#, 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 calculate the time it takes for the object to reach point B, you can use the kinematic equation for uniformly accelerated motion in three dimensions:

[ \vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2 ]

Where:

- ( \vec{r} ) is the final position vector (point B).
- ( \vec{r}_0 ) is the initial position vector (point A).
- ( \vec{v}_0 ) is the initial velocity vector (initially at rest, so ( \vec{v}_0 = 0 )).
- ( \vec{a} ) is the acceleration vector.
- ( t ) is the time taken.

Given:

- ( \vec{r}_0 = (7, 3, 9) )
- ( \vec{r} = (2, 5, 0) )
- ( \vec{a} = \left(0, 0, \frac{7}{4}\right) )

You can solve for ( t ) by substituting the known values into the equation and solving for ( t ).

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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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