Stopping Car Question Suppose that as a yellow car brakes, its velocity is described by $$v(t)=3.3e^{1-t}-0.6$$
If the brakes are applied at time t=0  seconds, what is the distance it takes for the car to come to a complete stop.
Round your answer to 3 decimal places. 
I first equalized this equation to zero to find at what time the car stops, and found $t=2.7047$.
After that, I made this $t$ the upper bound and $0$ the lower bound and integrated the equation, which I found $$-3.3e^{1-t}-0.6$$ as my integral.
I calculated the boundaries and found $8.3703$ as my final result, but looks like I am wrong. Can someone find my mistake? Thank you.
 A: The integral result is actually
$$d(t) = -3.3e^{1-t} - 0.6t + C \tag{1}\label{eq1}$$
The expression you give is $-3.3e^{1-t} - 0.6$, with the error coming from not the second term missing the factor of $t$. Using $t = 2.7047$ with your expression gives $d(2.7047) - d(0) = 3.3(e - e^{-1.7047})$, giving a result of $8.3683\ldots$. Your value of $8.3703$ comes from using a more accurate value of $t$, such as $t = 2.704748092238425234644711456507$ I get using Windows calculator. However, using these more accurate values, then subtracting $-0.6t$ (to get the equivalent of using \eqref{eq1}) gives $6.74748\ldots$. I trust this is the correct answer you were looking for.
A: First we calculate $v(t)=0$. To know where the car stops.
$3.3e^{1-t}-0.6=0\Leftrightarrow t=1-\ln(2/11)\approx 2.7047$
To get the distance, the car needed to stop, we have to calculate the following integral:
$\int_0^{1-\ln(2/11)} 3.3e^{1-t}-0.6\, dt=[-3.3e^{1-t}-0.6t]_0^{1-\ln(2/11)}$
$=(-3.3e^{\ln(2/11)}-0.6(1-\ln(2/11))-(-3.3e^{1}-0)=-3.3\cdot\frac{2}{11}-0.6+0.6\cdot\ln(2/11)+3.3e^1\approx 6.747$
