Will You Collide with the Car Ahead?

[negation]

Homework Statement

You're speeding at 85km/h when you notice that you're only 10m behind the car in front of you which is moving at the legal speed limit of 60km/h. You slam on your brakes and your car negatively accelerates at 4.2m/s/s. Assuming the other car continues at constant speed, will you collide? If so, at what relative speed?
Otherwise, what will be the distance between the cars at their closest approach?

Homework Equations

None.

The Attempt at a Solution

t = [vf-vi]/a
t = [0-23.61ms^-1]/-4.2ms^-2
t = 5.619s
at t = 5.619s, my car would have traveled 66m
the car moving at 60kmh^-1 would have traveled 93.65m
No collision occurs and the minimum distance between the 2 cars is (93.65m-66m)+10m

 

[BeBattey]

Can you list what equations you're using from your textbook in order to solve this?

 

[negation]

Can you list what equations you're using from your textbook in order to solve this?

No equations are given.

I've made another attempt at the solutions in my original post.

 

[BeBattey]

Well there are some equations that are always true under constant acceleration which should either be given to you or you should know enough calculus to derive them.

The one important here is: x(t)=x(0)+v(0)t+\frac{1}{2}at^{2}

This is an equation modeling the position of any object under constant acceleration. x(0) is the position when time is 0, and same with v(0).

If these model position for any time t, if you were to set up an equation modeling both cars positions, if you set them equal to you each other you would be saying that for some time, t, both cars occupy the same position, x.

So if that is solvable for a time, t, then they would have crashed. If it's unsolvable, then they never come into contact.

 

[BeBattey]

See this hyperphysics page: http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html

 

[negation]

Well there are some equations that are always true under constant acceleration which should either be given to you or you should know enough calculus to derive them.

The one important here is: x(t)=x(0)+v(0)t+\frac{1}{2}at^{2}

This is an equation modeling the position of any object under constant acceleration. x(0) is the position when time is 0, and same with v(0).

If these model position for any time t, if you were to set up an equation modeling both cars positions, if you set them equal to you each other you would be saying that for some time, t, both cars occupy the same position, x.

So if that is solvable for a time, t, then they would have crashed. If it's unsolvable, then they never come into contact.

The above kinematics equation brings me to the same answer as what I did too.

 

[BeBattey]

In your original post, you set vf=0, this would not be true. Your final velocity would be unknown at first glance. Could you show me your work using the equation I posted?

 

[negation]

In your original post, you set vf=0, this would not be true. Your final velocity would be unknown at first glance. Could you show me your work using the equation I posted?

x(5.619) = x(0) + 23.61ms^-1 (5.619) + 0.5(-4.2ms^-2)(5.619)^2 = 66.36m
(5.619) = x(0) + 16.67ms^-1(5.619) + 0.5(0)(5.619)^2 = 93.66m

(93.66-66.36) + 10

 

[BeBattey]

Show me your derivation of the time t at which their distance is minimized. I calculated a different t.

 

[negation]

Show me your derivation of the time t at which their distance is minimized. I calculated a different t.

t = [vf - vi]/a
t = [(0ms^-1) - 23.61ms^-1]/-4.2ms^-2
t = 5.619s

 

[BeBattey]

Ok, your vf would not be zero. This is your problem.

In collisions, you can tap each other but both remain in motion.

What you found was instead the time it would take for you to stop if you maintained the same rate of deceleration.

In order to find the minimum distance, you can construct a function OF their distance, namely x2-x1. I set mine equal to some arbitrary letter, let's say d, and then minimized it.

What do you know about minimization?

 

[negation]

Ok, your vf would not be zero. This is your problem.

In collisions, you can tap each other but both remain in motion.

In order to find the minimum distance, you can construct a function OF their distance, namely x2-x1. I set mine equal to some arbitrary letter, let's say d, and then minimized it.

What do you know about minimization?

Could you expound on "In order to find the minimum distance, you can construct a function OF their distance, namely x2-x1. I set mine equal to some arbitrary letter, let's say d, and then minimized it. "?
f">0?

 

[BeBattey]

Pretty much. The point where f''>0 and f'=0 is a local minimum.

The trick is in finding the function you're minimizing, the distance between the two cars as a function of time.

 

[negation]

Pretty much. The point where f''>0 and f'=0 is a local minimum.

The trick is in finding the function you're minimizing, the distance between the two cars as a function of time.

Could you provide the mathematical reasoning behind as to how I should set up the equation? I do have a more difficult time dealing with applied mathematics than pure mathematics.

 

[BeBattey]

The distance between any two points is one point minus the other point, at least on a one dimensional line.

So if those points are changing with time, the distance function also changes with time, but it always remains one point minus the other.

Just make those points functions of time.

 

[negation]

The distance between any two points is one point minus the other point, at least on a one dimensional line.

So if those points are changing with time, the distance function also changes with time, but it always remains one point minus the other.

Just make those points functions of time.

xf - xi = vit + 0.5at^2
d' = vi + at
-23.61ms^-1 = -4.2ms^-2 t
t = 5.6214

If I were to differentiate d', then wouldn't the time variable be non-existent? Can I deduce that since d" = a, then, this equation is unsolvable and therefore no collision occurs?

 

[BeBattey]

You have an equation for the position of both cars. If you were to subtract one from the other you could get a varying distance between the cars as a function of time. Note you're not using the initial and final velocities, you have a function of time.

Derive THAT equation and set it to 0.

 

[negation]

You have an equation for the position of both cars. If you were to subtract one from the other you could get a varying distance between the cars as a function of time. Note you're not using the initial and final velocities, you have a function of time.

Derive THAT equation and set it to 0.

EDIT:
d1 = vit + 0.5at^2
d2 = vit + 0.5at^2

d1 = 16.67ms^-1 t
d2 = 23.1,s^-1 t + 0.5(-4.2ms^-2)t^2

d2-d1 = 23.1ms^-1 t + 0.5(-4.2ms^-2)t^2 - 16.67ms^-1 t
d = 6.43ms^-1 t - 0.5(4.2ms^-2)t^2
t = 1.53

 

[BeBattey]

Alright, so you have a function that models position that change when time changes (it's a function of time)

This is: x(t)=x(0)+v(0)t+\frac{1}{2}at^2

So for each car we can construct their own, specific function of position:

x_{1}(t)=23.6t+\frac{1}{2}(-4.2)t^2
x_{2}(t)=10+16.7t

You've done this already to find if they would occupy the same point at any time t, so I don't feel like I'm doing the problem for you by outlining them now. You set them equal to each other, then noted you're not able to solve the equation, therefore no time t exists where they occupy the same point.

Now if we wanted to find the distance between the cars at any point, we'd calculate their positions with the above equations, then find their difference, say at time t=1s:

x_{1}(1)=23.6(1)+\frac{1}{2}(-4.2)(1)^2=21.5
x_{2}(1)=10+16.7(1)=26.7

So we know car 1 is at 21.5 and car 2 is at 26.7. Their distance between each other is then:

26.7-21.5=5.2

The key here is realizing that you don't need to evaluate the functions before you plug them into your typical difference between two points equation. Just leave them at some arbitrary time t, and set them equal to some function of distance between the two, say d(t):

d(t)=x_{2}(t)-x_{1}(t)

Then right there, you have a function of the distance between them as a function of time. You want to find when this function has a minimum value, so you apply what you know about minimizing functions and evaluate from there.

 

[negation]

Alright, so you have a function that models position that change when time changes (it's a function of time)

This is: x(t)=x(0)+v(0)t+\frac{1}{2}at^2

So for each car we can construct their own, specific function of position:

x_{1}(t)=23.6t+\frac{1}{2}(-4.2)t^2
x_{2}(t)=10+16.7t

You've done this already to find if they would occupy the same point at any time t, so I don't feel like I'm doing the problem for you by outlining them now. You set them equal to each other, then noted you're not able to solve the equation, therefore no time t exists where they occupy the same point.

Now if we wanted to find the distance between the cars at any point, we'd calculate their positions with the above equations, then find their difference, say at time t=1s:

x_{1}(1)=23.6(1)+\frac{1}{2}(-4.2)(1)^2=21.5
x_{2}(1)=10+16.7(1)=26.7

So we know car 1 is at 21.5 and car 2 is at 26.7. Their distance between each other is then:

26.7-21.5=5.2

The key here is realizing that you don't need to evaluate the functions before you plug them into your typical difference between two points equation. Just leave them at some arbitrary time t, and set them equal to some function of distance between the two, say d(t):

d(t)=x_{2}(t)-x_{1}(t)

Then right there, you have a function of the distance between them as a function of time. You want to find when this function has a minimum value, so you apply what you know about minimizing functions and evaluate from there.

so,
d(t) = 2.1t^2 + -6.9t + 10
d'(t) = 4.2t -6.9
t = 1.6s

d"(t) = 4.2

apparently, it's unsolvable.
Am I doing it right?

 

[BeBattey]

So ask yourself what's going on graphically.

You have a function of their distance, you found the point at which the slope is zero (first derivative is zero), and you also found that at that point where the first derivative is zero, the graph is concave up (second derivative is positive) so you know it's a local minimum of the graph.

If at time t=1.6s the distance between the two cars is a minimum, how do you find the actual distance between the two cars?

 

[negation]

So ask yourself what's going on graphically.

You have a function of their distance, you found the point at which the slope is zero (first derivative is zero), and you also found that at that point where the first derivative is zero, the graph is concave up (second derivative is positive) so you know it's a local minimum of the graph.

If at time t=1.6s the distance between the two cars is a minimum, how do you find the actual distance between the two cars?

Easy.

I would sub the time variable, t = 1.6s, into each of the car's respective distance as a function of time, x(t) = x(0) + v(0)t + 0.5at^2. Then find the difference between them.

 

[BeBattey]

Great!

Note you can also sub the time into the distance function you derived, because that's what you're looking for anyway.

Good job!