# An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #150 KJ# to # 42KJ# over #t in [0,8s]#. What is the average speed of the object?

The average speed is

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To find the average speed of the object, we need to use the formula for kinetic energy:

(KE = \frac{1}{2}mv^2)

Rearranging the formula to solve for velocity (v):

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

We'll first find the velocity at the initial kinetic energy and then at the final kinetic energy. Then, we'll calculate the average speed by finding the average of these velocities.

For the initial kinetic energy:

(KE_1 = 150 \text{ kJ} = 150 \times 10^3 \text{ J})

For the final kinetic energy:

(KE_2 = 42 \text{ kJ} = 42 \times 10^3 \text{ J})

Using the formula for velocity:

(v_1 = \sqrt{\frac{2 \times 150 \times 10^3}{5}} \approx 77.46 \text{ m/s})

(v_2 = \sqrt{\frac{2 \times 42 \times 10^3}{5}} \approx 49.29 \text{ m/s})

Now, we'll find the average speed:

(v_{\text{average}} = \frac{v_1 + v_2}{2} = \frac{77.46 \text{ m/s} + 49.29 \text{ m/s}}{2} \approx 63.38 \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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