One Dimensional Kinematics: Finding distance between two cars


by firekid123
Tags: 1 dimensional, acceleration, distance, kinematics
firekid123
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#1
Sep13-11, 07:02 PM
P: 8
1. The problem statement, all variables and given/known data

Car A is traveling a distance d behind Car B. Initically both cars are traveling at the same speed of 20.02 m/s. Suddenly Car B applies the brakes, causing Car B to decelerate at 3.05 m/s2. It takes the driver of Car A 0.75 seconds to react, and when she applies her brakes Car A decelerates at 3.658 m/s2.

What is the initial minimum distance d between the cars so as to avoid a collision?

2. Relevant equations

xf = xi + vix*t +(1/2)axt2

vfx = vix + ax*t

[itex]\Delta[/itex]x = vix*t -(1/2)axt2
3. The attempt at a solution

For Car B I found the final x position:
xf = 0m + 20.02m/s(.75s) +(1/2)(-3.05m/s)(.75s^2) = 14.157 m

And then because velocity for Car A in those .75 seconds is constant a=0. So:
[itex]\Delta[/itex]x = (20 m/s)(.75s) = 15 m

But then I'm not exactly sure what those 15m from Car A is.
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Delphi51
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#2
Sep13-11, 11:48 PM
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For Car B I found the final x position:
xf = 0m + 20.02m/s(.75s) +(1/2)(-3.05m/s)(.75s^2) = 14.157 m
It is a common error to use Δx = vix*t -(1/2)axt˛ in this way.
It is wrong because in the .75 s interval, the car is moving at constant speed and the formula does not apply. Just use Δx = vt for the first interval. To calculate the distance for the deceleration interval, you must first calculate its time.

This is quite a complicated bit of work because of the two cars. It would be a very good idea to sketch the distance and speed graphs just to keep track of things. You might even calculate the distances on the velocity graph - area under the graph for each car is its distance.
firekid123
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#3
Sep14-11, 01:24 AM
P: 8
The way I was setting everything up was that t = .75 is the time it takes for Car B to decelerate. Maybe that's where I'm confused and I even drew everything out. But I thought in the .75 seconds it took for Car A to eventually start decelerating, Car B was decelerating and at the end of .75 seconds it was stopped.

So Car B is not moving constant during t = .75? Or am I thinking of this wrong?

Here is more of what I did:


Car B: xf = 0m + 20.02m/s(.75s) +(1/2)(-3.05m/s)(.75s^2) = 14.157 m
Car A: [itex]\Delta[/itex]x = 20m/s(.75 s) + (1/2)(0)(.75^2) = 15 m. Because Car is is constant during t=.75 seconds the acceleration is 0.

Then I figured that the CarB - CarA = d -(14.157-15) And this is where I'm stuck. What do I do next?

Here is the diagram I made, please tell me if it's wrong: http://i.imgur.com/477B4.png

BruceW
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#4
Sep14-11, 08:18 AM
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One Dimensional Kinematics: Finding distance between two cars


I think Delphi51 mis-interpreted your equations. They are right. For car B, Δx is 14.157m and for car a, Δx is 15m (This is just during the time when B was decelerating and A was constant speed). So now the distance between the two cars is d - (15-14.157)m

What you need to do next is figure out how their distance changes during the time when they are both decelerating. You can do this by calculating the speeds of the two cars at the end of the last section and then working out how acceleration affects their positions.

Or you can just calculate the position of car B as a function of time, and the position of car A as a function of time (keeping in mind that it travelled at constant speed for a bit before decelerating). And then subtract these positions to get the distance between them, then solve for this distance to always be positive.

Or an easier way would be to think of the situation from the frame of reference where the cars are initially at rest. So from this perspective, car A is stationary for 0.75 seconds before it starts decelerating, and Car B will start decelerating at t=0, but it will have speed=0 at t=0, which makes the equations easier.
Delphi51
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#5
Sep14-11, 12:30 PM
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It takes the driver of Car A 0.75 seconds to react
means that car A continues in motion at constant speed 20.02 m/s for the 0.75 seconds.

It might be worth making a rough estimate. Car B is initially at 20 m/s, decelerating at 3 m/s˛, so it would take about 7 s to stop. For that time, its average speed is 10 m/s, so will travel distance 10x7 = 70 m.
firekid123
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#6
Sep14-11, 06:15 PM
P: 8
Quote Quote by BruceW View Post

What you need to do next is figure out how their distance changes during the time when they are both decelerating. You can do this by calculating the speeds of the two cars at the end of the last section and then working out how acceleration affects their positions.
If I were to set it up this way I'm not exactly sure what I'm supposed to be finding.

I solved
vfx = vix + axt for Car B:
20 + (-3.04)(.75) = 17.714 m/s but I'm not sure why I did this if that makes sense. I thought the final velocity would be 0 for car B because it's fully braked in the end.

And for car A I did solve it where the final velocity is 0 so I could find the time it took for it to decelerate and come to a stop.

0 = 20 m/s + (-3.658)t where t=5.467 So with t I would know the distance it traveled after the .75 seconds when it was going 20 m/s.

Then I used xf = xi +vixt + (1/2)axt2 for Car A:

0 = 15m + 20m/s(5.47s) + (1/2)(-3.658)(5.47)2 where xf= 69.67 m


I guess I'm confused now because I don't know what to do for Car B.
PeterO
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#7
Sep14-11, 08:01 PM
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Quote Quote by firekid123 View Post
If I were to set it up this way I'm not exactly sure what I'm supposed to be finding.

I guess I'm confused now because I don't know what to do for Car B.
Car B accelerates [negative acceleration if the initial velocity is positive] so you can use standard equations of motion to find how far it travels before stopping [as Delphi51 showed with the approximation, the answer should be around 70m.

Car A has two parts to its motion. a period where it travels at constant speed [covered the 15m you calculates] followed by a period where it accelerated. Giving the greater magnitude acceleration, it will take less distance to stop than B did,during the acceleration phase, but there is that extra 15m, so total distance will be greater - lets assume it is 82m

If Car B travel 70m while braking, and Car A travels 82m while reacting and braking, then Car a had better have been at least 12 m behind - d must be at least 12.

Once you have calculated correctly, you will have the real numbers, and thus get the real answer.

EDIT: I would have been using different equation of motion - the one that does not include time, because you don't know the time [v2 = u2 + 2ax is one form of that equation]
Delphi51
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#8
Sep14-11, 08:41 PM
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vfx = vix + axt for Car B:
20 + (-3.04)(.75)
This equation will also work, but you have the wrong time. It takes more than .75 seconds to stop. Use vf = vi + at to figure out t, with vf = 0.
firekid123
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#9
Sep14-11, 08:45 PM
P: 8
Alright I did all the exact numbers and I got 4.03 meters. Thanks for the help everyone!

Edit: I'll go change my numbers and then get the right amount.

EditEdit: I ignored using:
vfx = vix + axt for Car B:
20 + (-3.04)(.75), so it should still be right.

I think I was mixed up originally, I thought it took .75 seconds for Car B to decelerate and brake but that's not right.
mlotfi
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#10
Sep15-11, 02:13 PM
P: 2
how exactly did you get 4.03m? I understand the process up until this point. any help would be awesome!
firekid123
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#11
Sep15-11, 03:04 PM
P: 8
I went by what PeterO and Delphi51 said Correct me if I'm wrong with any steps:

Btime: 0 = 20.02 m/s + (-3.048 m/s^s)t
Btime = 6.56 s

delta x = 20.02 m/s (6.56 s) + (-3.048 m/s^s)(6.56s ^2)
= 65.75 m

for A:
delta x = 20.02 m/s (.75 s) + (0)(.75s^2)
deltax = 15 m
time: 0 = 20.02 m/s + (-3.658 m/s^2)t
time = 5.47 s

xf = 15m + 20.02m/s(5.47s) + (1/2)(-3.658 m/s^2)(5.47s ^2)
= 69.78 m

d must be at least 4.03 meters
PeterO
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#12
Sep15-11, 07:46 PM
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Quote Quote by firekid123 View Post
I went by what PeterO and Delphi51 said Correct me if I'm wrong with any steps:

Btime: 0 = 20.02 m/s + (-3.048 m/s^s)t
Btime = 6.56 s

delta x = 20.02 m/s (6.56 s) + (-3.048 m/s^s)(6.56s ^2)
= 65.75 m

for A:
delta x = 20.02 m/s (.75 s) + (0)(.75s^2)
deltax = 15 m
time: 0 = 20.02 m/s + (-3.658 m/s^2)t
time = 5.47 s

xf = 15m + 20.02m/s(5.47s) + (1/2)(-3.658 m/s^2)(5.47s ^2)
= 69.78 m

d must be at least 4.03 meters
Great work!
mlotfi
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#13
Sep15-11, 08:02 PM
P: 2
thank you very much, super clear explanation!


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