In a water- tank test involving the launching of a small model boat, the model's initial horizontal velocity is 6 m/s and its horizontal acceleration varies linearly from -12 m/s2 at t=0 to -2 m/s2 at t=t1 and then remains equal to -2 m/s2 until t=1.4 ?

Jun 24, 2015

I am not sure about what you need but I think it is ${t}_{1}$ or the velocity at this instant. I tried this but I am not sure, so have a look and check it!

Explanation:

The boat has a variable acceleration that can be described as linear from ${t}_{0} = 0$ up to ${t}_{1}$ and then constant up to ${t}_{2} = 1.4 s$.
Graphically:

Integrating the two expressions we can get to the velocity functions:
${v}_{a} \left(t\right) = \int \left(\frac{10}{t} _ 1 t - 12\right) \mathrm{dt} = \frac{5}{t} _ 1 {t}^{2} - 12 t + {c}_{1}$
setting at $t = 0 \to {v}_{a} \left(0\right) = 6 \frac{m}{s}$ we get: ${c}_{1} = 6$

so: $\textcolor{red}{{v}_{a} \left(t\right) = \frac{5}{t} _ 1 {t}^{2} - 12 t + 6}$

and:
${v}_{b} \left(t\right) = \int - 2 \mathrm{dt} = - 2 t + {c}_{2}$
setting at $t = 1.4 s \to {v}_{b} \left(1.4\right) = 0$ we get: ${c}_{2} = 2.8$ (notice that I supposed that the final velocity is zero!!! I am not sure about it!).

so: $\textcolor{red}{{v}_{b} \left(t\right) = - 2 t + 2.8}$

At $t = {t}_{1}$ the two curves should cross so:
$\frac{5}{t} _ 1 {t}_{1}^{2} - 12 {t}_{1} + 6 = - 2 {t}_{1} + 2.8$
$5 {t}_{1} - 12 {t}_{1} + 6 = - 2 {t}_{1} + 2.8$
$5 {t}_{1} = 3.2$
${t}_{1} = 0.64 s$
and $v \left(0.64\right) = 1.52 \frac{m}{s}$