Question

an object moving in a straight line travels s(t) kilometers in t hours, where s(t) = 9t² + 6t. (a) what is the objects velocity when t = 8? (b) how far has the object traveled in 8 hours? (c) when is the object traveling at a rate of 8 km/h?

Explanation:

Part (a)

Step1: Recall velocity as derivative of position

The velocity \( v(t) \) is the derivative of the position function \( s(t) \). For \( s(t) = 9t^2 + 6t \), we use the power rule: if \( f(t) = at^n \), then \( f'(t) = nat^{n - 1} \).
So, \( v(t) = s'(t) = \frac{d}{dt}(9t^2 + 6t) = 9\times2t + 6 = 18t + 6 \).

Step2: Substitute \( t = 8 \) into velocity function

Now, substitute \( t = 8 \) into \( v(t) \):
\( v(8) = 18\times8 + 6 = 144 + 6 = 150 \).

Step1: Recall distance is position function at \( t = 8 \)

To find the distance traveled in 8 hours, we evaluate the position function \( s(t) \) at \( t = 8 \).

Step2: Substitute \( t = 8 \) into \( s(t) \)

\( s(8) = 9\times(8)^2 + 6\times8 = 9\times64 + 48 = 576 + 48 = 624 \).

Step1: Set velocity equal to 8 and solve for \( t \)

We know \( v(t) = 18t + 6 \). Set \( v(t) = 8 \):
\( 18t + 6 = 8 \).

Step2: Solve the linear equation

Subtract 6 from both sides: \( 18t = 8 - 6 = 2 \).
Then divide both sides by 18: \( t = \frac{2}{18} = \frac{1}{9} \approx 0.111 \) hours.

Answer:

The object's velocity when \( t = 8 \) is \( 150 \) km/h.

Part (b)