Answer: Car A Crash Distance: Solving System of Equations

[SPhy]

Homework Statement

Car A is traveling at a speed of 31.0 m/s, car B is 3.5 meters behind car A traveling at the same speed. Suddenly, car A comes to a halting stop. Car B's reaction time to the stop is 0.85 seconds. Even if you assume both cars slow at the same rate, a crash will occur. Assume the magnitude of the braking acceleration is about 7.0 m/s^2 .

Determine the distance car A covers from when they apply the brakes to where they are rear ended by car B.

Text book hint: You may have to solve a system of equations.

Homework Equations

Car A: X(t) = (1/2)(a)t² + V_oAt + X_oA
Car B: X(t) = (1/2)(a)t² + V_oBt + X_oB

The Attempt at a Solution

Car B, during that time interval, has no acceleration. Also, I am going to assume the origin begins with car B. Therefore, car B's position function, during that time interval, simplifies to

X(t) = V_oBt = (31.0)(t)

Car A then has the position function

X(t) = -3.5t² + (31.0)t + 3.5

Setting these two functions equal to each-other yields

t = 1.0 seconds, so to find the distance car A has traveled to when they applied the brakes,

x(1) = 31 meters .

A big assumption I think I had to make was that Car B, did not have a lot of room to de-accelerate before crashing into car A.

Although I could have been totally off in my approach. In class we did a similar problem.

Thank you

 

[Simon Bridge]

A big assumption I think I had to make was that Car B, did not have a lot of room to de-accelerate before crashing into car A.

You are told how much room (= time) car B had to decelerate - don't assume.

It is best to do this sort of problem symbolically before you plug numbers in - it helps keep your reasoning clear.
It looks like you assumed the crash occurs inside the reaction time... i.e. car B does not decelerate at all. Is that correct? Check.

 

[SPhy]

You are told how much room (= time) car B had to decelerate - don't assume.

It is best to do this sort of problem symbolically before you plug numbers in - it helps keep your reasoning clear.
It looks like you assumed the crash occurs inside the reaction time... i.e. car B does not decelerate at all. Is that correct? Check.

Understood.

Hm, let's see, if we plug .85 seconds into car A's position function, we get, about 27.3 meters of travel. On the other hand, when we plug .85 seconds into car B's position function we get roughly 26.4 meters of travel. Which means, even after all the allotted reaction time has surpassed, there exists 1 meter of distance between the two cars. Which would then mean there is indeed room for car B to deaccelerate.

My guess would be I need to re-define car B's position function.

 

[SecretShooter]

Take a look at the initial distance between the two cars. then take a look at the deceleration rate and the reaction time.

 

[SPhy]

Take a look at the initial distance between the two cars. then take a look at the deceleration rate and the reaction time.

Is it possible for me to define the function, such that, car B's position function includes the deacceleration rate? Despite that, in the first .85 seconds, the speed is constant?

If so, perhaps I can write...

Car B

x(t) = -(1/2)(a)(t - .85)² + V_o(t - .85) + distance traveled after .85 seconds has passed

 

[SecretShooter]

Despite that, in the first .85 seconds, the speed is constant?

Yes.

So car A begins decelerating while car B does not. how long will they take to hit?

 

[Simon Bridge]

x(t) = -(1/2)(a)(t - .85)2 + Vo(t - .85) + distance traveled after .85 seconds has passed

... don't forget the position car B starts at.

It really helps to do it all symbolically.

i.e. Assuming car A breaks at x=0, when t=0...
$x_A(t)= ut-\frac{1}{2}at^2$ ... is your equation for how the position of car A changes with time.

At $t=0$ define: $x_0=x_A(0)-x_B(0) = 3.5\text{m}$ - is the initial separation of the cars.
So $x_B(0) = -x_0$ see?

If the reaction time is $t_r$, the equation for the position of car B is:
$x_B(t\leq t_r)=-x_0 + ut\\ x_B(t > t_r)=\; ?$

... you were quite close.

 

[SPhy]

Yes.

So car A begins decelerating while car B does not. how long will they take to hit?

Well, after .85 seconds, car B does have deceleration, and at that point, car A is still decelerating. Perhaps I can take advantage of the "reaction distance" of car B.

 

[SecretShooter]

Well, after .85 seconds, car B does have deceleration, and at that point, car A is still decelerating. Perhaps I can take advantage of the "reaction distance" of car B.

You haven't answered my question. I think when you do, the answer will be clear.

If both cars have an initial velocity of 31 m/s, and they are separated by 3.5 m, and car a begins decelerating at 7 m/s^2 how long will it take for the distance to be 0. If it's less than .85 seconds, then car B never even hits the brakes.

 

[SPhy]

You haven't answered my question. I think when you do, the answer will be clear.

If both cars have an initial velocity of 31 m/s, and they are separated by 3.5 m, and car a begins decelerating at 7 m/s^2 how long will it take for the distance to be 0. If it's less than .85 seconds, then car B never even hits the brakes.

Correct, but my difficulty had lied in finding that T value. Which I believe is roughly, 1.01 seconds. From that point I can plug into car A's position to find brake distance.

 

[SecretShooter]

You might want to check your math. I think you might be making it more complicated than it is. car B closes on car A at the rate of 7m/s^2 for .85 sec. how far is that?