The position (feet traveled) of a car is given by the equation s(t)=1/4t^2+1.
Find the time when the car is going the same speed as its average speed over the interval 0 to 10 seconds.
Here are the choices... my guess is "Never"
+9466 We need an equation that tells us the speed of the car at any time.
To do that, we can take the derivative of the position equation.
Let's call that function v(t) .
v(t) \(=\frac{d}{dt} (\frac14t^2+1) \\~\\ =\frac{d}{dt} \frac14t^2 +\frac{d}{dt}1 \\~\\ =(\frac14)(2)(t)+0 \\~\\ =\frac12t\)
And the average speed is given by...
average speed \(=\frac{\text{distance}}{\text{time}} \\~\\ =\frac{s(10)-s(0)}{10} \\~\\ =\frac{(\frac14(10)^2+1)-(\frac14(0)^2+1)}{10} \\~\\ =\frac{5}{2}\)
Now we just need to know what t equals when v(t) = \( \frac52 \)
\(\frac52 = \frac12t \\~\\ \frac{5}{2}\div\frac{1}{2} =t \\~\\ 5=t\)
