Minimum deceleration to prevent a collision

[Jenny Physics]

Homework Statement

At a given time $t=0$, a car 1 moving with constant velocity $v_{1}$ breaks with constant deceleration $d$. Another car 2 at a distance $L$ behind car 1 and traveling with constant acceleration $a_{2}$ and velocity $v_{2}$ at $t=0$ brakes immediately as soon as it sees car 1 brake (zero reaction time). Both cars brake with the same deceleration $d$. What is the critical (minimum) value of $d$ that will prevent the cars from colliding?

Homework Equations

Kinematics for velocity, acceleration, position

The Attempt at a Solution

[/B]
Car 1 has equation of motion $x_{1}=L+v_{1}t-\frac{1}{2}dt^{2}$.
Car 2 has equation of motion $x_{2}=v_{2}t-\frac{1}{2}dt^{2}$
(I don't think I need to use $a_{2}$?)

The cars will collide if $x_{1}=x_{2}$. But that doesn't give any constraint on $d$.

 

[haruspex]

The cars will collide if $x_{1}=x_{2}$. But that doesn't give any constraint on $d$.

Just looking at those equations, yes.
To see why, consider viewing one car from the frame of the other.
However, those two equations will not apply for all time. When will they cease to apply?

 

[kuruman]

Homework Statement

At a given time $t=0$, a car 1 moving with constant velocity $v_{1}$ breaks with constant deceleration $d$. Another car 2 at a distance $L$ behind car 1 and traveling with constant acceleration $a_{2}$ and velocity $v_{2}$ at $t=0$ brakes immediately as soon as it sees car 1 brake (zero reaction time). Both cars brake with the same deceleration $d$. What is the critical (minimum) value of $d$ that will prevent the cars from colliding?

Homework Equations

Kinematics for velocity, acceleration, position

The Attempt at a Solution

[/B]
Car 1 has equation of motion $x_{1}=L+v_{1}t-\frac{1}{2}dt^{2}$.
Car 2 has equation of motion $x_{2}=v_{2}t-\frac{1}{2}dt^{2}$
(I don't think I need to use $a_{2}$?)

The cars will collide if $x_{1}=x_{2}$. But that doesn't give any constraint on $d$.

If the cars start braking at the same time and have the same deceleration $d$, this means that their velocities will change in exactly the same way which also means that their relative velocity will not change, while the distance between them will keep decreasing. What must happen to avoid a collision?

This post has been edited to reflect a change of mind.

 

[Jenny Physics]

This problem doesn't make sense the way it is stated if we assume that the acceleration of each car changes discontinuously from $a_1$ and $a_2$ to deceleration $d$. If the cars start braking at the same time and have the same deceleration $d$, this means that their velocities will change in exactly the same way which also means that their relative velocity will not change. Thus, if $v_2 > v_1$, the cars will always collide regardless of the value of $d$. Are you sure you formulated the problem as was given to you?

Yes I think so. $a_{1}=0$ so I can see how the car goes from constant velocity to decreasing velocity suddenly. I think haruspex has a point that the equations might not apply all the time.

 

[Jenny Physics]

Just looking at those equations, yes.
To see why, consider viewing one car from the frame of the other.
However, those two equations will not apply for all time. When will they cease to apply?

They cease to apply when $x_{1}<0$ and/or $x_{2}<0$?

 

[kuruman]

They cease to apply when the acceleration of one car changes because they are valid only as long as the acceleration is constant. The leading car will come to rest first but the trailing car will still be moving. When the leading car stops, its acceleration changes from $d$ to zero.

I was careless in asserting that the cars will not collide regardless of $d$ when $v_2 > v_1$ in #3 and I edited the post to change that.

 

[PKM]

Car 1 has equation of motion $x_{1}=L+v_{1}t-\frac{1}{2}dt^{2}$.
Car 2 has equation of motion $x_{2}=v_{2}t-\frac{1}{2}dt^{2}$
(I don't think I need to use $a_{2}$?)

The cars will collide if $x_{1}=x_{2}$. But that doesn't give any constraint on $d$.

How can you take the time taken to decelerate $t$ be the same for two cars?
You should consider the eqn $v^2=u^2+2as$. It should help you to get $d$ in terms of $v_1,v_2$ & $L$.

Moreover, note that the critical condition arises when the car 2 stops at the exact point where car 1 stopped.

 

[haruspex]

How can you take the time taken to decelerate t be the same for two cars?

That's not how I read what @Jenny Physics did. She found the two displacements as functions of t and observed that the difference was independent of d. This is true so long as the accelerations are the same.

 

[PKM]

That's not how I read what @Jenny Physics did. She found the two displacements as functions of t and observed that the difference was independent of d. This is true so long as the accelerations are the same.

Yes, I get what she meant.
So long as both the cars keep moving the difference $x_2-x_1$ has nothing to do with $d$.
But I feel it is easier to avoid $t$ to solve the problem.

 

[haruspex]

it is easier to avoid t to solve the problem.

Yes, indeed.

 

[Jenny Physics]

How can you take the time taken to decelerate $t$ be the same for two cars?
You should consider the eqn $v^2=u^2+2as$. It should help you to get $d$ in terms of $v_1,v_2$ & $L$.

Moreover, note that the critical condition arises when the car 2 stops at the exact point where car 1 stopped.

The time it takes for the front car to stop is $v=v_{1}-dt=0$ or $t=\frac{v_{1}}{d}$.
At this time it will have traveled $x_{1}=L+\frac{v_{1}^{2}}{2d}$.
The car behind it will have traveled $x_{2}=\frac{v_{1}}{d}\left(v_{2}-\frac{v_{1}}{2}\right)$.
The minimum $d$ to avoid collision is when these distances equal each other which gives
$d=\frac{v_{1}(v_{2}-v_{1})}{L}$. For higher values of $d$, $x_{2}$ should be $<x_{1}$.

A question I have is why not consider the time it takes the car behind to stop instead of the car in front?

 

[kuruman]

Not quite.
You wrote $x_{1}=L+\frac{v_{1}^{2}}{2d}.$ In this equation $x_1$ is the position of the car when it stops. It is already distance $L$ ahead so, the distance it has traveled since it started braking is $\frac{v_{1}^{2}}{2d}$.

The car behind it will have traveled $x_{2}=\frac{v_{1}}{d}\left(v_{2}-\frac{v_{1}}{2}\right).$

How do you figure? To avoid the collision, the trailing car must come to rest after covering distance $L$ plus the distance traveled by the leading car.

 

[Jenny Physics]

Not quite.
You wrote $x_{1}=L+\frac{v_{1}^{2}}{2d}.$ In this equation $x_1$ is the position of the car when it stops. It is already distance $L$ ahead so, the distance it has traveled since it started braking is $\frac{v_{1}^{2}}{2d}$.

How do you figure? To avoid the collision, the trailing car must come to rest after covering distance $L$ plus the distance traveled by the leading car.

So I would have to find the time it takes for the back car to travel that distance which gives me the quadratic $v_{2}t-\frac{1}{2}dt^{2}=L+\frac{v_{1}^{2}}{2d}$. This gives a time $t=\frac{v_{2}+\sqrt{v_{2}^{2}+2dL+v_{1}^{2}}}{d}$. Then plugging that into the velocity of car 2 which must be zero so $v_{2}-dt=0$. This gives $0=\sqrt{v_{2}^{2}+2dL+2v_{1}^{2}}$ which is an impossible equation for $d$. So I have something wrong with the logic.

 

[PKM]

So I would have to find the time it takes for the back car to travel that distance which gives me the quadratic $v_{2}t-\frac{1}{2}dt^{2}=L+\frac{v_{1}^{2}}{2d}$. This gives a time $t=\frac{v_{2}+\sqrt{v_{2}^{2}+4dL+2v_{1}^{2}}}{2d}$. Then plugging that into the velocity of car 2 which must be zero so $v_{2}-dt=0$. This gives $v_{2}=\sqrt{v_{2}^{2}+4dL+2v_{1}^{2}}$ which is an impossible equation for $d$. So I have something wrong with the logic.

How can you equate the $t$ derived from the two eqns? Do the cars travel for the same time before coming to a halt?

 

[PKM]

Like I said earlier, consider the velocity-disp-accleration eqn, and set the values (you've all of them).

 

[Jenny Physics]

How can you equate the $t$ derived from the two eqns? Do the cars travel for the same time before coming to a halt?

Are you saying equate the time $t=\frac{v_{2}+\sqrt{v_{2}^{2}+2dL+v_{1}^{2}}}{d}$ to the time $2v_{1}^{2}/d$? The first is the time it takes the back car to reach the front car (although not with zero velocity). The second is the time it takes the front car to stop. Equating these 2 times gives me a value for $d$ but it doesn't guarantee that the back car stops there so it will collide with the car in front which is stopped.

 

[PKM]

Are you saying equate the time $t=\frac{v_{2}+\sqrt{v_{2}^{2}+2dL+v_{1}^{2}}}{d}$ to the time $2v_{1}^{2}/d$? The first is the time it takes the back car to reach the front car (although not with zero velocity). The second is the time it takes the front car to stop. Equating these 2 times gives me a value for $d$ but it doesn't guarantee that the back car stops there so it will collide with the car in front which is stopped.

I think the the eqn should be $t=\frac{v_{2}+\sqrt{v_{2}^{2}-2dL-v_{1}^{2}}}{d}$.
Yes, you can equate it to $t=\frac {v_2}{d}$. Sorry, I'd been wrong earlier, I'd missed an obvious point.

By the way, if you tried to use the eqn $v^2=u^2+2as$, it would be much more easier.

 

[Jenny Physics]

I think the the eqn yields $t=\frac{v_{2}+\sqrt{v_{2}^{2}-2dL-v_{1}^{2}}}{d}$.
Yes, you can equate it to $t=\frac {v_2}{d}$. Sorry, I'd been wrong earlier, I'd missed an obvious point.

By the way, if you tried to use the eqn $v^2=u^2+2as$, it would be much more easier.

You are right that it should be $t=\frac{v_{2}+\sqrt{v_{2}^{2}-2dL-v_{1}^{2}}}{d}$. Thank you. Equating to $\frac{v_{2}}{d}$ gives $d=\frac{v_{2}^{2}-v_{1}^{2}}{2L}$. I think I get it now. Both cars are at rest at this point. For higher values of $d$ however it is not intuitive to me that the back car will stop before it reaches the front car (since the back car has a higher velocity).

 

[PKM]

For higher values of $d$ however it is not intuitive to me that the back car will stop before it reaches the front car (since the back car has a higher velocity).

Yes, the back car will then hit the first one. Car 1 may or may not be in motion, but if $d$ exceeds this value, it is bound to be hit by car 2.
Is there any value of $d$ for which the back car hits the front one when it is still in motion? What do you feel?

 

[gneill]

Perhaps consider the stopping distance for each car. Start with the standard kinematic equation $v_f^2 - v_i^2 = 2 a s$. Obviously setting $V_f$ to zero. There's no time involved in this approach.

 

[Jenny Physics]

Yes, the back car will then hit the first one. Car 1 may or may not be in motion, but if $d$ exceeds this value, it is bound to be hit by car 2.
Is there any value of $d$ for which the back car hits the front one when it is still in motion? What do you feel?

Yes there will be many (most?) values of $d$ for which the two cars will collide. The question makes it look like above a certain $d$ the cars will not collide which I think it is false. There is probably just one $d$ for which that is true not a range of them.

 

[PKM]

The question makes it look like above a certain $d$ the cars will not collide which I think it is false. There is probably just one $d$ for which that is true not a range of them.

On the contrary, I feel if $d$ exceeds that value, the back car must hit the front one.
See it this way, say $d\lt \frac {{v_2}^2-{v_1}^2}{2L}$.
Then $(x_2-x_1)\gt L$, right?
That means car 2 necessarily hits car 1. (Here $x_1$ & $x_2$ are the total distance traveled by car 1 & 2 before coming to a halt)

In fact, we can also figure out whether the front car is still moving or not when the back car hits it (i.e. given $d\lt \frac {{v_2}^2-{v_1}^2}{2L}$).
Can you figure it out?

 

[Jenny Physics]

On the contrary, I feel if $d$ exceeds that value, the back car must hit the front one.
See it this way, say $d\gt \frac {{v_2}^2-{v_1}^2}{2L}$.
Then $(x_2-x_1)\gt L$, right?
That means car 2 necessarily hits car 1. (Here $x_1$ & $x_2$ are the total distance traveled by car 1 & 2 before coming to a halt)

In fact, we can also figure out whether the front car is still moving or not when the back car hits it (i.e. given $d\gt \frac {{v_2}^2-{v_1}^2}{2L}$).
Can you figure it out?

Yes I agree with you. What I mean is that there is only one $d$ for which the two cars will not collide.

 

[PKM]

Yes I agree with you. What I mean is that there is only one $d$ for which the two cars will not collide.

I had a typo there, check $d$ should be less than that value for the cars to collide, sorry.

But why do you feel there's only one value of $d$ for which the two cars will not collide? Check the inequalities and set $d$ greater than the critical value. You'll see that implies $(x_2-x_1)\lt L$. The cars will not collide then.

 

[Jenny Physics]

I had a typo there, check $d$ should be less than that value for the cars to collide, sorry.

But why do you feel there's only one value of $d$ for which the two cars will not collide? Check the inequalities and set $d$ greater than the critical value. You'll see that implies $(x_2-x_1)\lt L$. The cars will not collide then.

You had it right the first time. For $d\leq \frac{v_{2}^{2}-v_{1}^{2}}{2L}$ the two cars will not collide. Otherwise they will collide. This makes intuitive sense: The higher $d$ is the earlier the car in front stops and the less the car behind (with a higher velocity) has to catch up with the car in front.

 

[haruspex]

This makes intuitive sense

It may satisfy intuition but it is provably wrong.
It says that for a given d, L and v₁, collision is avoided by a sufficiently large v₂.

 

[Jenny Physics]

It may satisfy intuition but it is provably wrong.
It says that for a given d, L and v₁, collision is avoided by a sufficiently large v₂.

The front car stops when $t=\frac{v_{1}}{d}$ by which time it will be at position $x_{1}=L+\frac{v_{1}^{2}}{2d}$.
The car behind stops when $t=\frac{v_{2}}{d}$ by which time it will be at position $x_{2}=\frac{v_{2}^{2}}{2d}$.
Then $x_{2}-x_{1}=-L+\frac{v_{2}^{2}-v_{1}^{2}}{2d}$
If $d=a\frac{v_{2}^{2}-v_{1}^{2}}{2L},a>1$ then $x_{2}-x_{1}=L(a-1)>0$ i.e. $x_{2}>x_{1}$ and the cars collide.

 

[haruspex]

The front car stops when $t=\frac{v_{1}}{d}$ by which time it will be at position $x_{1}=L+\frac{v_{1}^{2}}{2d}$.
The car behind stops when $t=\frac{v_{2}}{d}$ by which time it will be at position $x_{2}=\frac{v_{2}^{2}}{2d}$.
Then $x_{2}-x_{1}=-L+\frac{v_{2}^{2}-v_{1}^{2}}{2d}$
If $d=a\frac{v_{2}^{2}-v_{1}^{2}}{2L},a>1$ then $x_{2}-x_{1}=L(a-1)>0$ i.e. $x_{2}>x_{1}$ and the cars collide.

Your (a-1) should be ((1/a)-1).

 

[Jenny Physics]

Your (a-1) should be ((1/a)-1).

Ha thank you!

 

[PKM]

By the way, we see that if $d$ is lesser than a certain value the back car will hit the front one. We still don't know if the front car is in motion on not when the back car hits it. Can you figure out the condition that the back car collides the front one, when front car is still moving?

 

[haruspex]

By the way, we see that if $d$ is lesser than a certain value the back car will hit the front one. We still don't know if the front car is in motion on not when the back car hits it. Can you figure out the condition that the back car collides the front one, when front car is still moving?

That goes back to post #1.

 

[PKM]

That goes back to post #1.

Why?
I actually meant to find the condition that the collision occurs before car 1 comes to rest. That wasn't asked in the question. The collision may occur when car 1 has already stopped, or still running. I asked about finding out a constraint to ensure which one occurs. Does it make sense now?

 

[haruspex]

Why?

Because the approach there assumed the two SUVAT equations quoted applied up to the point of collision. That was not valid for the original question but is true for the question you pose.

 

[Jenny Physics]

By the way, we see that if $d$ is lesser than a certain value the back car will hit the front one. We still don't know if the front car is in motion on not when the back car hits it. Can you figure out the condition that the back car collides the front one, when front car is still moving?

The time it takes for the front car to stop is $t_{1}=\frac{v_{1}}{2d}$. Assuming that both cars are decelerating the collision time is $t_{collision}=\frac{L}{v_{2}-v_{1}}$. This time is smaller or equal to $t_{1}$ if $d< \frac{v_{1}}{v_{1}+v_{2}}d_{critical}$ where $d_{critical}=\frac{v_{2}^{2}-v_{1}^{2}}{2L}$.
This is the range of $d$ for which the back car will hit the front car while the front car is still moving. This means that for $\frac{v_{1}}{v_{1}+v_{2}}d_{critical}\leq d<d_{critical}$ the back car will collide with a stopped front car.