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 ?

1 Answer
Jun 24, 2015

Answer:

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#