Completely stumped on this one -- Kinematic Conceptual problem

[ruskointhehizzy]

Homework Statement

You are at a stoplight when you see a car approaching from behind at constant velocity. To avoid getting rear-ended, you accelerate forward with constant acceleration. Assume that you have managed to start at the last possible instant to avoid getting hit, as determined by the oncoming car's speed and your car's acceleration. How far behind you is the other car when you begin to move?

known: no values are given.
variables: velocity and acceleration.

Homework Equations

x = vt, v = at, Vx = dr/dt where r is displacement.
ax = dVx/dt or d^2r/dt.
x velocity: Vx = V0x + axt
position: x = x0 + Vx0t + 1/2axt^2

The Attempt at a Solution

I was driving today and I looked in my rear view and thought about this, as the car comes closer it is speeding up, as it becomes more distant it is slowing down. Then I tried thinking about this problem and don't know how to think about this. Honestly I don't even know where to start, I asked about 10 other students in class and they didn't either. If I could have someone here ask me relevant questions that will help me think this through I would greatly appreciate that. Please do not just give me the answer but help me think it though and derive it - I want to understand the concepts being taught in this problem. Thank you so very much!

 

[ruskointhehizzy]

also this is the chapter before they introduce force, so I can only use the material from beforehand. We are using University Physics ed 14. and this is on ch3, which is about motion - so we can use vectors and the kinematic equations, but no stuff on force yet.
Thank you.

 

[haruspex]

don't know how to think about this

Just create some variables for the unknowns and see what equations you can write connecting them.
If you don't have enough equations, look for facts you have not used yet.
Post however far you get.

 

[CWatters]

+1 write some equations of motion for both. Then realize that for a crash to occur the two cars must be at the same place at the same time.

 

[Ray Vickson]

Homework Statement

You are at a stoplight when you see a car approaching from behind at constant velocity. To avoid getting rear-ended, you accelerate forward with constant acceleration. Assume that you have managed to start at the last possible instant to avoid getting hit, as determined by the oncoming car's speed and your car's acceleration. How far behind you is the other car when you begin to move?

known: no values are given.
variables: velocity and acceleration.

Homework Equations

x = vt, v = at, Vx = dr/dt where r is displacement.
ax = dVx/dt or d^2r/dt.
x velocity: Vx = V0x + axt
position: x = x0 + Vx0t + 1/2axt^2

The Attempt at a Solution

I was driving today and I looked in my rear view and thought about this, as the car comes closer it is speeding up, as it becomes more distant it is slowing down. Then I tried thinking about this problem and don't know how to think about this. Honestly I don't even know where to start, I asked about 10 other students in class and they didn't either. If I could have someone here ask me relevant questions that will help me think this through I would greatly appreciate that. Please do not just give me the answer but help me think it though and derive it - I want to understand the concepts being taught in this problem. Thank you so very much!

If you plot position $x$ vs. time $t$ for both the approaching car (call it A) and your car (call it Y), the graph $x=x_A(t)$ for A is a straight line of the form $x = -x_0+ vt,$ with slope $v$ equal to A's speed, while the graph $x=x_Y(t)$ for Y is a parabola of the form $x = \frac{1}{2} a t^2$. This assumes you start from rest and from the origin at time $t=0$, and car A is at $x=-x_0$ at that time.

In order to avoid a collision we need $x_Y(t) > x_A(t)$ for all $t \geq 0$, so either the Y-graph lies completely above the A-graph, or the A-graph is tangent to the Y-graph at a single point (corresponding to the cars just "touching" but not really colliding). Basically, you are being asked to find the value of $x_0$ in the A-graph.

 

[ruskointhehizzy]

If you plot position $x$ vs. time $t$ for both the approaching car (call it A) and your car (call it Y), the graph $x=x_A(t)$ for A is a straight line of the form $x = -x_0+ vt,$ with slope $v$ equal to A's speed, while the graph $x=x_Y(t)$ for Y is a parabola of the form $x = \frac{1}{2} a t^2$. This assumes you start from rest and from the origin at time $t=0$, and car A is at $x=-x_0$ at that time.

In order to avoid a collision we need $x_Y(t) > x_A(t)$ for all $t \geq 0$, so either the Y-graph lies completely above the A-graph, or the A-graph is tangent to the Y-graph at a single point (corresponding to the cars just "touching" but not really colliding). Basically, you are being asked to find the value of $x_0$ in the A-graph.

Okay I understand that the approaching car will be a straight line because it has constant speed, but I don't understand why X0 is negative? That totally makes sense to graph it I don't know why I didn't think of that. Could you explain to me how you got x = 1/2at^2? I can see how that will be a curve because you are going to have to floor it and you will gain speed at an exponential rate. I also don't understand what you mean by car A is at x = -x0 at t = 0.

Thank you so much. Here is a quick graph I made - is this correct?

 

[ruskointhehizzy]

thanks everyone

 

[Ray Vickson]

Okay I understand that the approaching car will be a straight line because it has constant speed, but I don't understand why X0 is negative? That totally makes sense to graph it I don't know why I didn't think of that. Could you explain to me how you got x = 1/2at^2? I can see how that will be a curve because you are going to have to floor it and you will gain speed at an exponential rate. I also don't understand what you mean by car A is at x = -x0 at t = 0.

Thank you so much. Here is a quick graph I made - is this correct?

I did not say $x_0 < 0$; in fact, we need $x_0 > 0$ because when you start from $x=0$ and accelerate to the right (to positive values of $x$), the approaching car is to your left on the x-axis, at position $-x_0 < 0$. (Alternatively, you could say the initial position of A is $x_0$, and require that we have $x_0 < 0$.)

And no, your graph is not correct. You have the cars together at time 0, just when you start to accelerate. Car A was supposed to be separated from you when you start to accelerate.

 

[haruspex]

you will gain speed at an exponential rate.

No, under constant acceleration you will gain speed at a uniform rate and distance at a quadratic rate.

 

[ruskointhehizzy]

I did not say $x_0 < 0$; in fact, we need $x_0 > 0$ because when you start from $x=0$ and accelerate to the right (to positive values of $x$), the approaching car is to your left on the x-axis, at position $-x_0 < 0$. (Alternatively, you could say the initial position of A is $x_0$, and require that we have $x_0 < 0$.)

And no, your graph is not correct. You have the cars together at time 0, just when you start to accelerate. Car A was supposed to be separated from you when you start to accelerate.

I guess I don't understand how you go the equation for the line A where you say it is in the form x = -$x_0$ + vt, because the equation in my book says x = $x_0$ + $v_{0x}$t - I see how $x_0$ will have to be greater then 0 now and I fixed my graph. I think I know what you are saying, it could either be like I have it or A could start at a -x value and Y would start at x = 0. Sorry man this may be trivial but I have only been doing physics for about 2 weeks and it is all still really new to me.

 

[ruskointhehizzy]

No, under constant acceleration you will gain speed at a uniform rate and distance at a quadratic rate.

right. Thanks for the correction.

 

[Ray Vickson]

I guess I don't understand how you go the equation for the line A where you say it is in the form x = -$x_0$ + vt, because the equation in my book says x = $x_0$ + $v_{0x}$t - I see how $x_0$ will have to be greater then 0 now and I fixed my graph. I think I know what you are saying, it could either be like I have it or A could start at a -x value and Y would start at x = 0. Sorry man this may be trivial but I have only been doing physics for about 2 weeks and it is all still really new to me.

OK, if you don't like $x = -x_0 + vt$ let's just change it to $x = c + vt$. At $t=0$ the position is $x=c$, and we had better have ##c < 0# because car A is not supposed to be at the position of car B yet.

Your graph is still not correct: car Y (you) starts off at speed 0, so the t-axis should be tangent to your (x,t)-graph at the starting point; you have the tangent sloping up, so somehow your car was instantly launched at a positive velocity---maybe on a catapult?

 

[ruskointhehizzy]

OK, if you don't like $x = -x_0 + vt$ let's just change it to $x = c + vt$. At $t=0$ the position is $x=c$, and we had better have ##c < 0# because car A is not supposed to be at the position of car B yet.

Your graph is still not correct: car Y (you) starts off at speed 0, so the t-axis should be tangent to your (x,t)-graph at the starting point; you have the tangent sloping up, so somehow your car was instantly launched at a positive velocity---maybe on a catapult?

I see what you mean lol - yes it was launched by a catapult, let me correct that.

 

[ruskointhehizzy]

thanks for sticking with me through this

 

[ruskointhehizzy]

I hope this graph is correct, I have the car Y sitting at the red light at x = 0 - and sees the oncoming car A at const v. Then before they hit he takes off - I think I captured that in this graph.

 

[haruspex]

Then before they hit he takes off

No. You have the car at the lights waiting until collision is just about to occur, then instantly moving at speed v, which would require infinite acceleration.
Forget the question for the moment and just graph displacement of a car undergoing constant acceleration from rest.

 

[ruskointhehizzy]

oooh

 

[ruskointhehizzy]

here is the displacement for an object with constant acceleration. on the right is showing a line with zero acceleration and the effect that the acceleration has on it.

 

[ruskointhehizzy]

I have now just added the A line to that.

 

[Ray Vickson]

I have now just added the A line to that.

You are getting close. You have assumed that YOU are at $x_0 > 0$ when you begin accelerating. Of course, you are free to consider your starting point to be anywhere you want, so maybe you do start from $x = x_0 = 147.3$. However, that seems too complicated. Life is much simpler if you assume you start from $x = 0$, with YOUR $x_0 = 0$ but the approaching car's $x_0 < 0$.

The reason I am belaboring this issue is that I am not convinced you really "get it". You should think about it and proceed carefully. Drawings can be rough sketches, but they should try to capture the main features of a problem.

Again, let me emphasize that you can choose starting points however you want, so you could start from $x = p$ at the moment the approaching car is at $x = q$. What you really want is to determine the distance $p-q$.

 

[ruskointhehizzy]

You have assumed that YOU are at x0>0x_0 > 0 when you begin accelerating. Of course, you are free to consider your starting point to be anywhere you want, so maybe you do start from x=x0=147.3x = x_0 = 147.3. However, that seems too complicated. Life is much simpler if you assume you start from x=0x = 0, with YOUR x0=0x_0 = 0 but the approaching car's x0<0x_0 < 0.

that makes total sense.

you could start from $x = p$ at the moment the approaching car is at $x = q$. What you really want is to determine the distance $p-q$.

right - that make sense.

Okay I will work on this more, thank you. I think I have a good picture of what is going on but you are right I don't fully understand it, which I want to make sure I do.

 

[PrathameshR]

 

[haruspex]

View attachment 213004

Looks right.
You can get there a bit faster by using the frame of reference of one car and running time backwards. With constant acceleration a, when the distance reaches d the speed difference is v.

 

[Hendrik Boom]

You need to decide what the origin of your graph is. Is it whatever you are at time zero? Is it where the other car is? And remember. You start at zero velocity, so your line will not start out at an angle.

Sent from my Nexus 5 using Physics Forums mobile app

 

[ruskointhehizzy]

View attachment 213004

thanks man - I went through it and I got this:

I thought about it - and I noticed you d is facing the opposite way of s1 and s2, is d negative and thus why you switched the inequality sign? Thanks again that problem kicked my ass but seeing it solved really helps. In math I am used to getting values and solving problems but in something like this where they don't give me any values really just gave me writers block, if that is a thing in math/physics :P. Take care.

 

[haruspex]

View attachment 213004

When I replied earlier I failed to notice that you were not the original poster of the thread.
Forum rules: do not post complete solutions for others' homework problems. The idea is to point out mistakes, clear up misunderstandings and give hints, that is all.

 

[PrathameshR]

thanks man - I went through it and I got this:

I thought about it - and I noticed you d is facing the opposite way of s1 and s2, is d negative and thus why you switched the inequality sign? Thanks again that problem kicked my ass but seeing it solved really helps. In math I am used to getting values and solving problems but in something like this where they don't give me any values really just gave me writers block, if that is a thing in math/physics :P. Take care.

I was getting close to end of my board so skipped some steps hence flipped inequality by mistake. d distance and not position or displacement so there's no need to label it as positive or negative. Sorry for not providing the hints and directly stating the answer.

 

[PrathameshR]

When I replied earlier I failed to notice that you were not the original poster of the thread.
Forum rules: do not post complete solutions for others' homework problems. The idea is to point out mistakes, clear up misunderstandings and give hints, that is all.

I'm sorry. I'm new to forum. It'll take some time to learn all rules

 

[ruskointhehizzy]

When I replied earlier I failed to notice that you were not the original poster of the thread.
Forum rules: do not post complete solutions for others' homework problems. The idea is to point out mistakes, clear up misunderstandings and give hints, that is all.

I was going to mention that but it was due yesterday so seeing the answer I feel is okay since I won't get credit for it anyways.

I was getting close to end of my board so skipped some steps hence flipped inequality by mistake. d distance and not position or displacement so there's no need to label it as positive or negative. Sorry for not providing the hints and directly stating the answer.

that's fine man I was able to work through it - I appreciate the solution, it helped me learn in this situation but I learn best when asked questions so I can provide my own solution. I know your intentions are best at heart though, and this was due yesterday so it is totally fine. Thank you again man. No worries on the rules I am new here as well man. Take care!

 

[DoctorPhysics]

Okay I understand that the approaching car will be a straight line because it has constant speed, but I don't understand why X0 is negative? That totally makes sense to graph it I don't know why I didn't think of that. Could you explain to me how you got x = 1/2at^2? I can see how that will be a curve because you are going to have to floor it and you will gain speed at an exponential rate. I also don't understand what you mean by car A is at x = -x0 at t = 0.

Thank you so much. Here is a quick graph I made - is this correct?

I'm just going off an assumption here. Could it be because Xo is prior to your starting point? So anything g behind your initial position is negative?

 

[Ray Vickson]

I'm just going off an assumption here. Could it be because Xo is prior to your starting point? So anything g behind your initial position is negative?

Of course. Surely you must have seen a "number line" before? East = positive $x$, west = negative $x$.

 

[kuruman]

Now that the problem has been marked "solved", I can contribute a quicker way to get to the answer. The required distance $x_{min.}$ will depend on two independent parameters, the constant speed $v_0$ of the trailing car and the maximum acceleration $a$ of the leading car. The leading car must reach speed $v_0$ over the minimum distance $x_{min.}$. If it moves slower than that, it started accelerating too late; if it moves faster than that, it started accelerating too soon. Therefore, from the "fourth" kinematic equation $$2ax_{min.}=v_0^2 - 0^2 \rightarrow x_{min.}=\frac{v_0^2}{2a}$$

 

[haruspex]

Now that the problem has been marked "solved", I can contribute a quicker way to get to the answer. The required distance $x_{min.}$ will depend on two independent parameters, the constant speed $v_0$ of the trailing car and the maximum acceleration $a$ of the leading car. The leading car must reach speed $v_0$ over the minimum distance $x_{min.}$. If it moves slower than that, it started accelerating too late; if it moves faster than that, it started accelerating too soon. Therefore, from the "fourth" kinematic equation $$2ax_{min.}=v_0^2 - 0^2 \rightarrow x_{min.}=\frac{v_0^2}{2a}$$

C.f. post #23.

 

[kuruman]

C.f. post #23.

Sure, but it's not necessary to run time backwards. In either car's reference frame the other car drops its speed from initial speed $v_0$ to zero over distance $x_{min.}$.

 

[haruspex]

Sure, but it's not necessary to run time backwards. In either car's reference frame the other car drops its speed from initial speed $v_0$ to zero over distance $x_{min.}$.

Yes, it is not necessary to run time backwards; I was just trying to make it exactly like the commonest example of the equation.