# 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

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

DoctorPhysics
Ray Vickson
Homework Helper
Dearly Missed

## 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

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.

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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?

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ruskointhehizzy
thanks everyone

Ray Vickson
Homework Helper
Dearly Missed
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
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Gold Member
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.

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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
Homework Helper
Dearly Missed
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.

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haruspex
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Gold Member
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.

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ruskointhehizzy
I have now just added the A line to that.

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Ray Vickson
Homework Helper
Dearly Missed
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.

haruspex
Homework Helper
Gold Member
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.

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
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.

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