The displacement of a particle moving along an `x` axis is given by `x=18t+5.0t^(2)`, where `x` is in meters and t is in seconds. Calculate (a) the instantaneous velocity at `t=2.0s` and (b) the average velocity between `t=2.0s` and `t=3.0s`.

Answer

To solve the problem, we will follow these steps:

### Given:
The displacement of a particle is given by the equation:
\[ x(t) = 18t + 5.0t^2 \]
where \( x \) is in meters and \( t \) is in seconds.

### (a) Calculate the instantaneous velocity at \( t = 2.0 \, \text{s} \)

1. **Differentiate the displacement function**:
The instantaneous velocity \( v(t) \) is the derivative of the displacement \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(18t + 5.0t^2) \]

2. **Apply the differentiation**:
Using the power rule of differentiation:
\[ v(t) = 18 + 10t \]

3. **Substitute \( t = 2.0 \, \text{s} \)**:
Now, we will find the instantaneous velocity at \( t = 2.0 \):
\[ v(2) = 18 + 10(2) = 18 + 20 = 38 \, \text{m/s} \]

### (b) Calculate the average velocity between \( t = 2.0 \, \text{s} \) and \( t = 3.0 \, \text{s} \)

1. **Find the displacement at \( t = 2.0 \, \text{s} \)**:
\[ x(2) = 18(2) + 5(2^2) = 36 + 20 = 56 \, \text{m} \]

2. **Find the displacement at \( t = 3.0 \, \text{s} \)**:
\[ x(3) = 18(3) + 5(3^2) = 54 + 45 = 99 \, \text{m} \]

3. **Calculate the change in displacement (\( \Delta x \))**:
\[ \Delta x = x(3) - x(2) = 99 - 56 = 43 \, \text{m} \]

4. **Calculate the change in time (\( \Delta t \))**:
\[ \Delta t = 3.0 - 2.0 = 1.0 \, \text{s} \]

5. **Calculate the average velocity**:
The average velocity \( v_{avg} \) is given by:
\[ v_{avg} = \frac{\Delta x}{\Delta t} = \frac{43 \, \text{m}}{1.0 \, \text{s}} = 43 \, \text{m/s} \]

### Final Answers:
(a) The instantaneous velocity at \( t = 2.0 \, \text{s} \) is \( 38 \, \text{m/s} \).

(b) The average velocity between \( t = 2.0 \, \text{s} \) and \( t = 3.0 \, \text{s} \) is \( 43 \, \text{m/s} \).

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