# An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #96 KJ# to # 280 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

The average speed is

The initial kinetic energy is

So,

The final kinetic energy is

So,

On a graph velocity v/s time, we have a straight line

Area of

The average speed

so,

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To find the average speed of the object, we need to calculate the change in kinetic energy and then use it to find the average speed.

The change in kinetic energy is the final kinetic energy minus the initial kinetic energy: ( \Delta KE = KE_{\text{final}} - KE_{\text{initial}} ).

Given that the initial kinetic energy is 96 kJ and the final kinetic energy is 280 kJ, the change in kinetic energy is:

( \Delta KE = 280 , \text{kJ} - 96 , \text{kJ} = 184 , \text{kJ} ).

Now, we know that the change in kinetic energy is equal to the work done on the object, which is given by the formula:

( \Delta KE = \frac{1}{2}mv^2 ), where ( m ) is the mass of the object and ( v ) is its final velocity.

Rearranging the formula to solve for ( v ):

( v = \sqrt{\frac{2\Delta KE}{m}} ).

Plugging in the values, we get:

( v = \sqrt{\frac{2 \times 184 , \text{kJ}}{4 , \text{kg}}} ).

( v = \sqrt{\frac{368}{4}} ).

( v = \sqrt{92} ).

( v ≈ 9.59 , \text{m/s} ).

Therefore, the average speed of the object is approximately ( 9.59 , \text{m/s} ).

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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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