# An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #72 KJ# to # 720 KJ# over #t in [0, 3 s]#. What is the average speed of the object?

The average speed is

The kinetic energy is

Therefore,

and,

The equation of the line is

So,

So,

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The average speed of the object is 8 m/s.

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The average speed of the object can be calculated using the formula:

[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} ]

Since the object's kinetic energy uniformly changes, we can assume that its acceleration is constant. Using the equation for kinetic energy:

[ KE = \frac{1}{2} m v^2 ]

where ( KE ) is the kinetic energy, ( m ) is the mass of the object, and ( v ) is the speed of the object.

Given that the initial kinetic energy is 72 kJ and the final kinetic energy is 720 kJ, we can find the initial and final speeds using these equations:

[ \text{Initial kinetic energy:} \quad 72 = \frac{1}{2} \times 6 \times v_{\text{initial}}^2 ]

[ \text{Final kinetic energy:} \quad 720 = \frac{1}{2} \times 6 \times v_{\text{final}}^2 ]

Solving for ( v_{\text{initial}} ) and ( v_{\text{final}} ), we get:

[ v_{\text{initial}} = \sqrt{\frac{72 \times 2}{6}} ]

[ v_{\text{final}} = \sqrt{\frac{720 \times 2}{6}} ]

The average speed is then:

[ \text{Average speed} = \frac{v_{\text{initial}} + v_{\text{final}}}{2} ]

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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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