For the previous question, the acceleration of the particle at any time `t` is :

Options

A. ` -0.8 m//s^(2)`
B. `0.8 m//s^(2)`
C. `-0.6 m//s^(2)`
D. `0.5 m//s^(2)`

Explanation

To find the acceleration of the particle at any time \( t \), we start with the given displacement function:

\[ s(t) = 15t - 0.4t^2 \]

### Step 1: Differentiate the displacement to find velocity

The first step is to differentiate the displacement function \( s(t) \) with respect to time \( t \) to find the velocity \( v(t) \).

\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(15t - 0.4t^2) \]

Using the power rule of differentiation:

\[ v(t) = 15 - 0.8t \]

### Step 2: Differentiate the velocity to find acceleration

Next, we differentiate the velocity function \( v(t) \) with respect to time \( t \) to find the acceleration \( a(t) \).

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(15 - 0.8t) \]

Again, applying the power rule:

\[ a(t) = 0 - 0.8 = -0.8 \, \text{m/s}^2 \]

### Conclusion

The acceleration of the particle at any time \( t \) is:

\[ a(t) = -0.8 \, \text{m/s}^2 \]

This indicates that the particle is experiencing a constant retardation.

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