# An object is at rest at #(7 ,8 ,4 )# and constantly accelerates at a rate of #5/4 m/s^2# as it moves to point B. If point B is at #(1 ,5 ,3 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

It takes 3.29 seconds.

First of all we need to know how much distance the objects has to travel. The distance between two points is given by the Pitagora's theorem in 3 dimensions

in our case

Now that we know that our body will travel for 6.78 m we can use the equation of motion for an accelerating body that is

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Using the kinematic equation for uniformly accelerated motion:

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

[ \Delta y = v_{yi} t + \frac{1}{2} a t^2 ]

[ \Delta z = v_{zi} t + \frac{1}{2} a t^2 ]

Where: ( \Delta x = x_B - x_A = 1 - 7 = -6 ) m, ( \Delta y = y_B - y_A = 5 - 8 = -3 ) m, ( \Delta z = z_B - z_A = 3 - 4 = -1 ) m, ( v_{xi} = 0 ) m/s, ( v_{yi} = 0 ) m/s, ( v_{zi} = 0 ) m/s, ( a = \frac{5}{4} ) m/s².

Substituting the values into the equations:

[ -6 = 0 + \frac{1}{2} \left( \frac{5}{4} \right) t^2 ]

[ -3 = 0 + \frac{1}{2} \left( \frac{5}{4} \right) t^2 ]

[ -1 = 0 + \frac{1}{2} \left( \frac{5}{4} \right) t^2 ]

Solving for ( t ):

[ -6 = \frac{5}{8} t^2 ] [ -3 = \frac{5}{8} t^2 ] [ -1 = \frac{5}{8} t^2 ]

[ t^2 = \frac{-6 \times 8}{5} = -\frac{48}{5} ] [ t^2 = \frac{-3 \times 8}{5} = -\frac{24}{5} ] [ t^2 = \frac{-1 \times 8}{5} = -\frac{8}{5} ]

Since time cannot be negative, the object will never reach point B. There may be an error in the calculations or assumptions.

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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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