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 ?

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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 ?

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Answer 1 · 2015-06-24T13:12:59

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 $t_2=1.4s$ .
Graphically:
enter image source here
Integrating the two expressions we can get to the velocity functions:
$v_a(t)=int(10/t_1t-12)dt=5/t_1t^2-12t+c_1$
setting at $t=0->v_a(0)=6m/s$ we get: $c_1=6$

so: $color(red)(v_a(t)=5/t_1t^2-12t+6)$

and:
$v_b(t)=int-2dt=-2t+c_2$
setting at $t=1.4s->v_b(1.4)=0$ we get: $c_2=2.8$ (notice that I supposed that the final velocity is zero!!! I am not sure about it!).

so: $color(red)(v_b(t)=-2t+2.8)$

At $t=t_1$ the two curves should cross so:
$5/t_1t_1^2-12t_1+6=-2t_1+2.8$
$5t_1-12t_1+6=-2t_1+2.8$
$5t_1=3.2$
$t_1=0.64s$
and $v(0.64)=1.52m/s$

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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 ?

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Explanation:

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