A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation $x(t)= αt^2− βt^3$, where$ α = 1.45 m/s^2$ and $β = 0.055 m/s^3$
What is the avg velocity from $t=0$ to $t=4.1$?
Deriving $x(t)$ is get $v(t) = 2αt-3βt^2$. I get $v(4.1) = 9.11635 m/s$
Plugging this in to find avg velocity from t =0 to t=4.1 I get: $\frac {(9.11635-0)}{(4.1-0)}$ = 2.2235 m/s
Is this correct? I am entering this into an online program and it is telling me it is wrong even though I answered a previous question with the same formula except from t = 0 to t = 1.9s. It accepted the previous question's answer with no problem.
We have \begin{align} \text{average velocity}&=\frac{x(4.1)-x(0)}{4.1-0\text{ s}}\\ &=\frac{(1.45\text{ m/s}^2)(4.1\text{ s})^2-(0.055\text{ m/s}^3)(4.1\text{ s})^3-0}{4.1\text{ s}}\\ &\approx \color{blue}{5.02\text{ m/s}} \end{align}