Change in Acceleration so a = 0 and v = 0
Since velocity is the antiderivative of acceleration $int{a(t)}dt$. Would we be able to calculate the change in acceleration needed such that acceleration and velocity would equal zero at the same time?
calculus
asked Nov 26 at 15:27
Danny
12
add a comment |
Since velocity is the antiderivative of acceleration $int{a(t)}dt$. Would we be able to calculate the change in acceleration needed such that acceleration and velocity would equal zero at the same time?
calculus
asked Nov 26 at 15:27
Danny
12
You can graph this situation easily, by drawing your velocity curve tangent to y=0 with a minimum/maximum at y=0. This would mean that v(t)=0 where y=0, and since it is a min or max a(t)=0.
– Zach
Nov 26 at 16:34
add a comment |
Since velocity is the antiderivative of acceleration $int{a(t)}dt$. Would we be able to calculate the change in acceleration needed such that acceleration and velocity would equal zero at the same time?
calculus
asked Nov 26 at 15:27
Danny
12
Since velocity is the antiderivative of acceleration $int{a(t)}dt$. Would we be able to calculate the change in acceleration needed such that acceleration and velocity would equal zero at the same time?
calculus
calculus
asked Nov 26 at 15:27
Danny
12
asked Nov 26 at 15:27
Danny
12
edited Nov 26 at 15:35
asked Nov 26 at 15:27
Danny
12
asked Nov 26 at 15:27
Danny
12
asked Nov 26 at 15:27
Danny
12
12
You can graph this situation easily, by drawing your velocity curve tangent to y=0 with a minimum/maximum at y=0. This would mean that v(t)=0 where y=0, and since it is a min or max a(t)=0.
– Zach
Nov 26 at 16:34
add a comment |
You can graph this situation easily, by drawing your velocity curve tangent to y=0 with a minimum/maximum at y=0. This would mean that v(t)=0 where y=0, and since it is a min or max a(t)=0.
– Zach
Nov 26 at 16:34
You can graph this situation easily, by drawing your velocity curve tangent to y=0 with a minimum/maximum at y=0. This would mean that v(t)=0 where y=0, and since it is a min or max a(t)=0.
– Zach
Nov 26 at 16:34
You can graph this situation easily, by drawing your velocity curve tangent to y=0 with a minimum/maximum at y=0. This would mean that v(t)=0 where y=0, and since it is a min or max a(t)=0.
– Zach
Nov 26 at 16:34
add a comment |
1 Answer
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active
oldest
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One needs to start with a function for acceleration that goes to $0$. I assume that at $t=0$ the velocity is $v_0>0$. At the same instant, one starts breaking, so the acceleration is negative. The simplest model is linear dependence of acceleration, but you can use whatever function you want, if you follow the same steps. $$a=-a_0+alpha t$$
Here $a_0,alpha>0$. The velocity is given by $$v(t)=v_0+int_0^ta(tau)dtau$$
So when velocity is $0$ we have $$v_0-a_0t+frac 12 alpha t^2=0$$
You also have from the acceleration that $$-a_0+alpha t=0$$
You have now two equations and three variables ($t,a_0,alpha$), so you would need to add one more constraint. That is up to you. You can say for example that the maximum deceleration is $a_0$, or you can specify that the vehicle should stop in a certain amount of time.
answered Nov 26 at 16:35
Andrei
11k21025
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
One needs to start with a function for acceleration that goes to $0$. I assume that at $t=0$ the velocity is $v_0>0$. At the same instant, one starts breaking, so the acceleration is negative. The simplest model is linear dependence of acceleration, but you can use whatever function you want, if you follow the same steps. $$a=-a_0+alpha t$$
Here $a_0,alpha>0$. The velocity is given by $$v(t)=v_0+int_0^ta(tau)dtau$$
So when velocity is $0$ we have $$v_0-a_0t+frac 12 alpha t^2=0$$
You also have from the acceleration that $$-a_0+alpha t=0$$
You have now two equations and three variables ($t,a_0,alpha$), so you would need to add one more constraint. That is up to you. You can say for example that the maximum deceleration is $a_0$, or you can specify that the vehicle should stop in a certain amount of time.
answered Nov 26 at 16:35
Andrei
11k21025
add a comment |
One needs to start with a function for acceleration that goes to $0$. I assume that at $t=0$ the velocity is $v_0>0$. At the same instant, one starts breaking, so the acceleration is negative. The simplest model is linear dependence of acceleration, but you can use whatever function you want, if you follow the same steps. $$a=-a_0+alpha t$$
Here $a_0,alpha>0$. The velocity is given by $$v(t)=v_0+int_0^ta(tau)dtau$$
So when velocity is $0$ we have $$v_0-a_0t+frac 12 alpha t^2=0$$
You also have from the acceleration that $$-a_0+alpha t=0$$
You have now two equations and three variables ($t,a_0,alpha$), so you would need to add one more constraint. That is up to you. You can say for example that the maximum deceleration is $a_0$, or you can specify that the vehicle should stop in a certain amount of time.
answered Nov 26 at 16:35
Andrei
11k21025
add a comment |
One needs to start with a function for acceleration that goes to $0$. I assume that at $t=0$ the velocity is $v_0>0$. At the same instant, one starts breaking, so the acceleration is negative. The simplest model is linear dependence of acceleration, but you can use whatever function you want, if you follow the same steps. $$a=-a_0+alpha t$$
Here $a_0,alpha>0$. The velocity is given by $$v(t)=v_0+int_0^ta(tau)dtau$$
So when velocity is $0$ we have $$v_0-a_0t+frac 12 alpha t^2=0$$
You also have from the acceleration that $$-a_0+alpha t=0$$
You have now two equations and three variables ($t,a_0,alpha$), so you would need to add one more constraint. That is up to you. You can say for example that the maximum deceleration is $a_0$, or you can specify that the vehicle should stop in a certain amount of time.
answered Nov 26 at 16:35
Andrei
11k21025
One needs to start with a function for acceleration that goes to $0$. I assume that at $t=0$ the velocity is $v_0>0$. At the same instant, one starts breaking, so the acceleration is negative. The simplest model is linear dependence of acceleration, but you can use whatever function you want, if you follow the same steps. $$a=-a_0+alpha t$$
Here $a_0,alpha>0$. The velocity is given by $$v(t)=v_0+int_0^ta(tau)dtau$$
So when velocity is $0$ we have $$v_0-a_0t+frac 12 alpha t^2=0$$
You also have from the acceleration that $$-a_0+alpha t=0$$
You have now two equations and three variables ($t,a_0,alpha$), so you would need to add one more constraint. That is up to you. You can say for example that the maximum deceleration is $a_0$, or you can specify that the vehicle should stop in a certain amount of time.
answered Nov 26 at 16:35
Andrei
11k21025
answered Nov 26 at 16:35
Andrei
11k21025
answered Nov 26 at 16:35
Andrei
11k21025
answered Nov 26 at 16:35
Andrei
11k21025
11k21025
add a comment |
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You can graph this situation easily, by drawing your velocity curve tangent to y=0 with a minimum/maximum at y=0. This would mean that v(t)=0 where y=0, and since it is a min or max a(t)=0.
– Zach
Nov 26 at 16:34