Question

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 ?

Answer 1

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:

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`