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Imagine the following scenario:

We have 2 carts moving alone a track, initially at x velocity. They are spaced apart by y metres and are both z length.

When a cart passes a pre-determined point on the track, it starts to decelerate down to another velocity. If the deceleration value is too high (i.e. the cart decelerates too quickly), a collision may occur as the back of the cart could still be before the deceleration point, thus causing the second cart to collide with it.

How could we determine the most appropriate decel rate which doesn't cause a collision?

I'm not on any maths or physics course, and in all likelihood (assuming I haven't massively over complicated this) I'm not going to be familiar with what's need to solve the problem. If somebody could point me in the right direction, in terms of equation, etc, that would be extremely helpful and allow me to do my own research.

M-R
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  • Feel free to throw keywords at me... – M-R Jul 31 '16 at 00:26
  • Normally you consider the cart to be rigid, so the whole cart decelerates at once. The point would be that the rear cart doesn't reach the deceleration point at the same time as the first, so continues at the original velocity, catching up to the leading cart. – Ross Millikan Jul 31 '16 at 00:26
  • @RossMillikan, yes the cart is rigid. Can you equip me with the necessary equations, etc? Thanks – M-R Jul 31 '16 at 00:28

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Let the initial velocity be $v$ (rather than $x$) and set the zero so the rear of the first cart is at $x=0$ when the deceleration starts and the zero of time when the deceleration starts. The acceleration is $-a$, where the $-$ sign shows the deceleration. We decelerate to a velocity $w$. The speed of the first cart is then $v_1=\begin {cases} v-at&t \lt \frac {v-w}{a}\\w&t \ge \frac {v-w}{a} \end {cases}$ We integrate that to get the position of the rear of the train $x_1=\begin {cases} vt-\frac 12at^2&t \lt \frac {v-w}{a}\\\frac {v(v-w)}a-\frac12a(\frac {v-w}{a})^2+wt&t \ge \frac {v-w}{a} \end {cases}$. The front of the second cart starts at $-y$, so its speed is $v_2=\begin {cases} v&t \lt \frac yv\\v-at&\frac yv \lt t \lt \frac yv+\frac {v-w}{a}\\w&t \ge \frac yv+\frac {v-w}{a} \end {cases}$ Again you integrate this to get the position $x_2=\begin {cases} -y+vt& t \lt \frac {y+z}v\\v(t-\frac {y+z}v)-\frac 12a(t-\frac {y+z}v)^2&\frac {y+z}v \lt t \lt \frac {y+z}v+\frac {v-w}{a}\\\frac {v(v-w)}a-\frac12a(\frac {v-w}{a})^2+w(t-\frac {y+z}v)&t \ge \frac {v-w}{a}+\frac {y+z}v \end {cases}$
where I used the fact that the time to start the deceleration of the second cart is $\frac {y+z}v$ as the front of the second cart has to travel $y+z$ before coming to the deceleration point. Now all you have to do is see if $x_2 \gt x_1$ at the end of the deceleration of the second cart. After that they are traveling the same speed.

Ross Millikan
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  • Could you explain what the curly brackets mean? thanks – M-R Jul 31 '16 at 01:24
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    The curly braces mean a case statement. The first says that when $t \lt \frac {v-w}a$ then $v_1=v-at$ and at other times $v_1=w$. This comes from the fact that the velocity decreases linearly until $v_1=w$ and stays constant after that. – Ross Millikan Jul 31 '16 at 02:46
  • Ahhh got it! One other thing: I've noticed that your solution simply tells you whether the two carts have collided during the deceleration. How would you calculate the highest possible deceleration-rate before a collision would occur? – M-R Jul 31 '16 at 10:50
  • For a given deceleration rate, there will be a total distance the rear cart gains on the front one. You can find that be comparing $x_1-x_2$ for $t$ large with $y$. If you do that analytically, you can see how it depends on $a$ – Ross Millikan Jul 31 '16 at 14:10