# An object is at rest at #(3 ,5 ,1 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(7 ,9 ,2 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

The time taken is

The following equation of motion will be used:

Thus,

In this case,

Thus,

Utilizing the motion equation,

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To find the time it takes for the object to reach point B, we can use the formula for displacement in uniformly accelerated motion:

[ \Delta x = v_0 t + \frac{1}{2} a t^2 ]

Where:

- ( \Delta x ) is the displacement (change in position),
- ( v_0 ) is the initial velocity (which is zero since the object is at rest),
- ( a ) is the acceleration,
- ( t ) is the time.

Given:

- Initial position (( x_0, y_0, z_0 )) = (3, 5, 1) meters
- Final position (( x_f, y_f, z_f )) = (7, 9, 2) meters
- Acceleration (( a )) = ( \frac{4}{3} ) m/s²

Substituting the given values into the formula and solving for ( t ):

[ 7 - 3 = 0 \cdot t + \frac{1}{2} \cdot \frac{4}{3} \cdot t^2 ] [ 4 = \frac{2}{3} \cdot t^2 ] [ t^2 = \frac{4 \cdot 3}{2} ] [ t^2 = 6 ] [ t = \sqrt{6} ]

Thus, it will take the object ( \sqrt{6} ) seconds 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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- The position of an object moving along a line is given by #p(t) = 3t - tcos(( pi )/3t) #. What is the speed of the object at #t = 5 #?
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