# An object is at rest at #(9 ,7 ,5 )# and constantly accelerates at a rate of #5/4 m/s^2# as it moves to point B. If point B is at #(8 ,2 ,6 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

The time is

We utilize the equation of motion.

so,

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Using the kinematic equation ( \Delta x = v_i t + \frac{1}{2} a t^2 ), where ( \Delta x ) is the displacement, ( v_i ) is the initial velocity, ( a ) is the acceleration, and ( t ) is the time, we can find the time it takes for the object to reach point B.

Given: Initial position ( \vec{r}_i = (9, 7, 5) ) Final position ( \vec{r}_f = (8, 2, 6) ) Acceleration ( a = 5/4 , \text{m/s}^2 ) Initial velocity ( v_i = 0 ) (since the object is at rest initially)

To find the displacement in each direction: ( \Delta x = r_{f_x} - r_{i_x} = 8 - 9 = -1 ) m ( \Delta y = r_{f_y} - r_{i_y} = 2 - 7 = -5 ) m ( \Delta z = r_{f_z} - r_{i_z} = 6 - 5 = 1 ) m

Using the formula ( \Delta x = v_i t + \frac{1}{2} a t^2 ) for each direction: For ( x )-direction: ( -1 = 0 + \frac{1}{2} \left(\frac{5}{4}\right) t^2 ) For ( y )-direction: ( -5 = 0 + \frac{1}{2} \left(\frac{5}{4}\right) t^2 ) For ( z )-direction: ( 1 = 0 + \frac{1}{2} \left(\frac{5}{4}\right) t^2 )

Solving each equation for ( t ): For ( x )-direction: ( t^2 = \frac{-2}{\frac{5}{4}} = -\frac{8}{5} ), which is not possible since time cannot be negative. For ( y )-direction: ( t^2 = \frac{-10}{\frac{5}{4}} = -\frac{40}{5} = -8 ), which is not possible since time cannot be negative. For ( z )-direction: ( t^2 = \frac{2}{\frac{5}{4}} = \frac{8}{5} ) ( t = \sqrt{\frac{8}{5}} = \frac{2\sqrt{2}}{\sqrt{5}} = \frac{2\sqrt{10}}{5} ) seconds

Therefore, it will take ( \frac{2\sqrt{10}}{5} ) 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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