Car collision with constant deceleration

[Alexstre]

I think I solved this whole thing but it seems a bit too simple. I think I might've missed a detail somewhere. This is the problem, followed by my steps:

Homework Statement

Two cars are about hit each other (face to face). The first car is doing 20m/s and decelerating at a rate of 5m/s^2 while the second car is doing 15m/s decelerating at a rate of 7.5m/s^2. When the drivers realize their situation the cars are 25 meters apart.

1. Find the moment (after how many second) will the crash happen.
2. What distance did the first car do before hitting the second car?
3. What is the speed of both cars as they hit each other?

The Attempt at a Solution

1. Find the moment (after how many second) will the crash happen.
Car A
Info: 20m/s, deceleration 5m/s^2 (a=-5m/s^2)
Function: x=xi+(vi*t)+0.5(at^2)
xi = 0, vi = 20m/s, t is time, a is -5m/s^2 so I simplified it to:
x = 20t - 2.5t^2

Car B
Info: 15m/s, deceleration 7.5m/s^2 (a=-7.5m/s^2)
Function: x=xi+(vi*t)+0.5(at^2)
xi = 0, vi = 15m/s, t is time, a is -7.5m/s^2 so I simplified it to:
x = 15t - 3.75t^2

What I did after is simply plug in 't' (0, 1, 2...) into each function until their sum was greater than 25m (the distance that initially separated both vehicles). I found that after 1 second:

xa(1) = 20(1) - 2.5(1^2) = 17.5 meters
xb(1) = 15(1) - 3.75(1^2) = 11.25 meters
Total: 28.75 meters, which is greater than 25, the cars hit each other a bit before 1 second.

This seem a bit too easy... but I kept going anyways.

2. What distance did the first car do before hitting the second car?
I got this from the first question. 17.5 meters. This is where I started realizing my first part might've been wrong. I was expecting the problem to be a bit more precise. Yes, if the car drives for 17.5 meters it will hit the second car, but it actually hits it BEFORE that so 17.5 meters isn't a precise answer here.

3. What is the speed of both cars as they hit each other?
I found dx/dt of my first functions:
va(t) = 20 - 5t
vb(t) = 15 - 7.5t

Now that makes sense. The initial speed of each car, minus the deceleration rate * the amount of second they slowed down for.

speed of A: 15 m/s
speed of B: 7.5 m/s

Anyone notices if I did something wrong? I'm thinking maybe the functions I used may not be exactly what I think they are, but going through my notes I can't find anything more appropriate for the problem!

Thanks a lot!

 

[BishopUser]

"the cars hit each other before 1 second" that is a pretty general statement, but yes. Of course you want to know exactly when they hit though.

Your equation is x=xi+(vi*t)+0.5(at^2). Why not set this equation up for each car using the same coordinate system? Using your method, you set the equation up with separate coordinate systems - this is not very helpful.

If you set up the equation for each car using the same coordinate system being very careful to assign negative and positive values appropriately, I have this feeling you end up with two equations with two unknowns and being able to solve for the variable that you like.

 

[tiny-tim]

Hi Alexstre!

Hint: use the relative acceleration and the relative distance, to find the time.

 

[Alexstre]

"the cars hit each other before 1 second" that is a pretty general statement, but yes. Of course you want to know exactly when they hit though.

Your equation is x=xi+(vi*t)+0.5(at^2). Why not set this equation up for each car using the same coordinate system? Using your method, you set the equation up with separate coordinate systems - this is not very helpful.

If you set up the equation for each car using the same coordinate system being very careful to assign negative and positive values appropriately, I have this feeling you end up with two equations with two unknowns and being able to solve for the variable that you like.

There is a value of t for which:
25 - (20t+2.5(t^2) - 15t+3.75(t^2)) = 0
That is, 25 meters between cars - (total distance traveled by both cars) = 0

That value of t should be a bit more precise. I think? :)

Hi Alexstre!

Hint: use the relative acceleration and the relative distance, to find the time.

I think that comes up to having one of the car going towards the right (assuming a 1 dimension movement) and the other one to the right. The car heading left has a negative speed and a positive acceleration (relative to the plan). Makes sense now!

Thanks!

 

[Jokerhelper]

There is a value of t for which:
25 - (20t+2.5(t^2) - 15t+3.75(t^2)) = 0
That is, 25 meters between cars - (total distance traveled by both cars) = 0

That value of t should be a bit more precise. I think? :)


I think that comes up to having one of the car going towards the right (assuming a 1 dimension movement) and the other one to the right. The car heading left has a negative speed and a positive acceleration (relative to the plan). Makes sense now!

Thanks!

First I have a question: have you ever studied relative motion or Galilean transformation? If no, that's fine. your approach is correct, but if you are going to add up vectors then it's fundamental that you set up a point of reference, such as an origin point of a coordinate system (which in this case will be linear as you are dealing with one dimension). If let's say you choose the initial position of car A to be your origin, then car B will have to have an initial displacement x_i that you will have to take into account for that equation.