- #1
VinnyCee
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The Problem:
A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48ft. above the ground. If the elevator can accelerate at [itex]0.6 \frac{ft.}{s^2}[/itex], decelerate at [itex]0.3 \frac{ft.}{s^2}[/itex], and reach a maximum speed of [itex]8 \frac{ft.}{s}[/itex], determine the shortest time to make the lift, starting from rest and ending at rest.
Work:
[tex]s\,=\,s_0\,+\,v_0\,t\,+\,\frac{1}{2}\,a_c\,t_2[/tex]
[tex]s_2\,=\,\left(0\right)\,+\,\left(0\right)\,\left(t_1\right)\,+\,\frac{1}{2}\,\left(0.6\,\frac{ft.}{s^2}\right)\,\left(t_1^2\right)[/tex]
[tex]48\,ft.\,=\,\left(s_2\right)\,+\,\left(8\frac{ft.}{s}\right)\,\left(t_2\right)\,+\,\frac{1}{2}\,\left(-0.3\,\frac{ft.}{s^2}\right)\,\left(t_1^2\right)[/tex]
[tex]s_2\,=\,0.15t_2^2\,-\,8t_2\,+\,48[/tex]
[tex]0.3\,t_1^2\,=\,0.15\,t_2^2\,-\,8\,t_2\,+\,48[/tex]
Here is my list of variables, starting with the final ones and going down to the initial ones:
[tex]v_3\,=\,0\,\,&\,\,s_3\,=\,48\,ft.[/tex]
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[tex]t_2\,=\,?[/tex]
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[tex]v_2\,=\,8\,\frac{ft.}{s}\,\,&\,\,s_2\,=\,?[/tex]
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[tex]t_1\,=\,?[/tex]
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[tex]v_1\,=\,0\,\,&\,\,s_1\,=\,0[/tex]
When I use other constant acceleration equations to solve for a position (s), I get an impossible answer like 53.3 ft. when the elevator is only supposed to go to 40 ft. at max! Please help, thanks in advance.
A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48ft. above the ground. If the elevator can accelerate at [itex]0.6 \frac{ft.}{s^2}[/itex], decelerate at [itex]0.3 \frac{ft.}{s^2}[/itex], and reach a maximum speed of [itex]8 \frac{ft.}{s}[/itex], determine the shortest time to make the lift, starting from rest and ending at rest.
Work:
[tex]s\,=\,s_0\,+\,v_0\,t\,+\,\frac{1}{2}\,a_c\,t_2[/tex]
[tex]s_2\,=\,\left(0\right)\,+\,\left(0\right)\,\left(t_1\right)\,+\,\frac{1}{2}\,\left(0.6\,\frac{ft.}{s^2}\right)\,\left(t_1^2\right)[/tex]
[tex]48\,ft.\,=\,\left(s_2\right)\,+\,\left(8\frac{ft.}{s}\right)\,\left(t_2\right)\,+\,\frac{1}{2}\,\left(-0.3\,\frac{ft.}{s^2}\right)\,\left(t_1^2\right)[/tex]
[tex]s_2\,=\,0.15t_2^2\,-\,8t_2\,+\,48[/tex]
[tex]0.3\,t_1^2\,=\,0.15\,t_2^2\,-\,8\,t_2\,+\,48[/tex]
Here is my list of variables, starting with the final ones and going down to the initial ones:
[tex]v_3\,=\,0\,\,&\,\,s_3\,=\,48\,ft.[/tex]
|
|
[tex]t_2\,=\,?[/tex]
|
|
[tex]v_2\,=\,8\,\frac{ft.}{s}\,\,&\,\,s_2\,=\,?[/tex]
|
|
[tex]t_1\,=\,?[/tex]
|
|
[tex]v_1\,=\,0\,\,&\,\,s_1\,=\,0[/tex]
When I use other constant acceleration equations to solve for a position (s), I get an impossible answer like 53.3 ft. when the elevator is only supposed to go to 40 ft. at max! Please help, thanks in advance.
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