Questions: An object is moving with velocity (in ft / sec) v(t) = t^2 - 5t + 4. At t=0, the object has a position of s=8 feet. Use antiderivatives to find the position equation. s(t) = The position of the object at 6 seconds is feet.

Solution Steps

The velocity function is given by \( v(t) = t^2 - 5t + 4 \). To find the position function \( s(t) \), we need to find the antiderivative of the velocity function.

The antiderivative of \( v(t) \) is:

\[ s(t) = \int (t^2 - 5t + 4) \, dt \]

Calculate the antiderivative term by term:

• The antiderivative of \( t^2 \) is \( \frac{t^3}{3} \).
• The antiderivative of \( -5t \) is \( -\frac{5t^2}{2} \).
• The antiderivative of \( 4 \) is \( 4t \).

Thus, the antiderivative is:

\[ s(t) = \frac{t^3}{3} - \frac{5t^2}{2} + 4t + C \]

We know that at \( t = 0 \), the position \( s = 8 \) feet. Substitute these values into the position equation to solve for \( C \):

\[ 8 = \frac{0^3}{3} - \frac{5 \cdot 0^2}{2} + 4 \cdot 0 + C \]

\[ 8 = C \]

Thus, the constant of integration \( C \) is 8.

Substitute \( C = 8 \) back into the position function:

\[ s(t) = \frac{t^3}{3} - \frac{5t^2}{2} + 4t + 8 \]

Substitute \( t = 6 \) into the position function to find the position at 6 seconds:

\[ s(6) = \frac{6^3}{3} - \frac{5 \cdot 6^2}{2} + 4 \cdot 6 + 8 \]

Calculate each term:

• \( \frac{6^3}{3} = \frac{216}{3} = 72 \)
• \( \frac{5 \cdot 6^2}{2} = \frac{5 \cdot 36}{2} = \frac{180}{2} = 90 \)
• \( 4 \cdot 6 = 24 \)

Thus:

\[ s(6) = 72 - 90 + 24 + 8 = 14 \]

Final Answer