To determine the instantaneous velocity of the race car at \( t = 4 , \text{s} \), we need to find the derivative of the position function \( x(t) \) with respect to time \( t \).
The position function is given as:
\[ x(t) = (5 , \text{m/s}) t + (0.75 , \text{m/s}) t^3 \]
Now, we differentiate \( x(t) \) to find the velocity function \( v(t) \):
\[ v(t) = \frac{dx(t)}{dt} = \frac{d}{dt} \left( (5 , \text{m/s}) t + (0.75 , \text{m/s}) t^3 \right) \]
Using the power rule, we differentiate each term:
- The derivative of \( (5 , \text{m/s}) t \) is \( 5 , \text{m/s} \).
- The derivative of \( (0.75 , \text{m/s}) t^3 \) is \( 3 \cdot (0.75 , \text{m/s}) t^2 = (2.25 , \text{m/s}) t^2 \).
Combining these results:
\[ v(t) = 5 , \text{m/s} + (2.25 , \text{m/s}) t^2 \]
Next, we substitute \( t = 4 , \text{s} \) into the velocity function to find the instantaneous velocity:
\[ v(4) = 5 , \text{m/s} + (2.25 , \text{m/s}) (4 , \text{s})^2 \]
Calculating the \( t^2 \) term:
\[ (4 , \text{s})^2 = 16 , \text{s}^2 \]
Thus:
\[ v(4) = 5 , \text{m/s} + (2.25 , \text{m/s}) \cdot 16 , \text{s}^2 \] \[ v(4) = 5 , \text{m/s} + 36 , \text{m/s} \] \[ v(4) = 41 , \text{m/s} \]
Therefore, the instantaneous velocity at \( t = 4 , \text{s} \) is \( \boxed{41 , \text{m/s}} \).
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