# Question #8ecac

Oct 11, 2017

Well, let's takes $f ' \left(t\right) = 10 {e}^{1.05} t = 10 {e}^{\frac{20}{21} t}$

If $y = a {e}^{b x}$, where $a$ and $b$ are constants, then $\frac{\mathrm{dy}}{\mathrm{dx}} = a b {e}^{b x}$.

So, $10 {e}^{1.05 t} = a b {e}^{b t}$.

Therefore, $a \cdot 1.05 = 10$.

$a = \frac{10}{1.05} = \frac{200}{21}$

So, $\int f \left(v\right) \mathrm{dt} = \frac{200 {e}^{1.05 t}}{21}$

This will be rewritten as ${v}_{y} = \frac{200 {e}^{1.05 t}}{21}$

This will now be rearranged to get $t = \ln \frac{\frac{21 {v}_{y}}{200}}{1.05}$

To find the total time taken, you do $\ln \frac{\frac{21 {v}_{2}}{200}}{1.05} - \ln \frac{\frac{21 {v}_{1}}{200}}{1.05}$. Where ${v}_{2}$ is the final velocity, and ${v}_{1}$ is the initial velocity.

This is all I can give you since I don't have the full question, only what you gave which was a part 4 to a, and b.

If your value for $t$ is negative, just make it positive by taking the value of $\left\mid t \right\mid$