# An object is at rest at #(8 ,2 ,9 )# and constantly accelerates at a rate of #3 m/s# as it moves to point B. If point B is at #(6 ,7 ,5 )#, 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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Find the shortest distance between point A and B, then using calculus, use the definition of acceleration and velocity to solve for the displacement in terms of time from the given acceleration, then solve for time using the previously calculated shortest distance, to get

But first, let's talk about the initial displacement. We can see that, from point A to B, the object has moved:

Now, we just need the time, but we may have to play around with acceleration and velocity for that. We don't yet know the velocity, and we can take the integral of acceleration to get that...

It may feel weird that I already have it set up for all three variables even though the Pythagorean theorem is originally for two variables. However, it really is just a result of attempting to apply the Pythagorean theorem twice:

So, using the definition of acceleration:

Then, what is the definition of velocity?

Again, the constant is just zero:

Interesting, we can now take the square root:

That's a werid amount of seconds, and it's certainly an irrational number... we could approximate it using a calculator:

There we go! That's what our calculations tell us about the time it takes for an object with the given acceleration and initial state to travel through the two given points.

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To calculate 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 0 since the object is at rest),
- ( a ) is the acceleration (given as 3 m/s²), and
- ( t ) is the time.

Given that the initial velocity (( u )) is 0 and the displacement (( s )) is the distance between the initial and final points (given as (\sqrt{(6-8)^2 + (7-2)^2 + (5-9)^2})), we can rearrange the equation to solve for ( t ):

[ t = \sqrt{\frac{2s}{a}} ]

Substituting the values, we get:

[ t = \sqrt{\frac{2 \times \sqrt{(6-8)^2 + (7-2)^2 + (5-9)^2}}{3}} ]

Calculating the distance and solving for ( t ), we find:

[ t ≈ \sqrt{\frac{2 \times \sqrt{13^2 + 5^2 + (-4)^2}}{3}} ≈ \sqrt{\frac{2 \times \sqrt{170}}{3}} ≈ \sqrt{\frac{2 \times 13}{3}} ≈ \sqrt{\frac{26}{3}} ≈ \frac{\sqrt{78}}{3} ]

[ t ≈ \frac{\sqrt{78}}{3} ]

Therefore, it will take approximately ( \frac{\sqrt{78}}{3} ) 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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