# If an object with uniform acceleration (or deceleration) has a speed of #4 m/s# at #t=2# and moves a total of 25 m by #t=6#, what was the object's rate of acceleration?

The acceleration is

Apply the equation of motion

The acceleration is

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To find the object's rate of acceleration, we can use the following formula:

[v_f = v_i + at]

Given:

- (v_i = 4 , \text{m/s}) (initial velocity)
- (v_f = ?) (final velocity)
- (t = 6 - 2 = 4 , \text{s}) (time interval)
- (d = 25 , \text{m}) (total distance traveled)

Using the formula:

[v_f = v_i + at]

[v_f = 4 , \text{m/s} + a(4 , \text{s})]

[v_f = 4 , \text{m/s} + 4a]

Now, we also know that:

[d = v_i t + \frac{1}{2}at^2]

Plugging in the given values:

[25 , \text{m} = (4 , \text{m/s})(4 , \text{s}) + \frac{1}{2}a(4 , \text{s})^2]

[25 , \text{m} = 16 , \text{m} + 8a]

[8a = 25 , \text{m} - 16 , \text{m}]

[8a = 9 , \text{m}]

[a = \frac{9 , \text{m}}{8 , \text{s}^2}]

[a = 1.125 , \text{m/s}^2]

So, the object's rate of acceleration is (1.125 , \text{m/s}^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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