When does the crash happen?

In summary, the conversation revolves around trying to calculate the time at which a crash between a car and a truck can be prevented. The first person is attempting to solve the problem by assuming that the truck stops until its velocity is 0.1, while the second person is questioning what assumptions are being made and whether there is a specific condition for the crash not to happen. The end result is that the solution does not seem to make sense and further clarification is needed.
  • #1
bolzano95
89
7
Homework Statement
A car is driving at the distance ##d= 15m## after the truck. Both vehicles drive with the constant velocity. Velocity of the car is ## v_1= 30,6 \frac{m}{s}##, velocity of the truck is ## v_2= 25 \frac{m}{s}##. Suddenly the truck starts to slow down with deceleration ##-a=5\frac{m}{s^2}##.
1. Calculate the time necessary so that the car and the truck don't crash. You know the car driver starts slowing down (##0,5s##) later than the truck and both the truck and the car decelerate with ##-a=5\frac{m}{s^2}##.
2. Calculate the time of the crash.
Relevant Equations
Basic kinematic equations
1. I'm trying to calculate the time at which the crash does not happen (if possible, because I don't know the official solution. I assume the crash is preventable).
At the time t the truck decelerates and makes the distance ##s_2= \frac{v_2^2}{2a}##. In the same time the car drives with the constant velocity ##v_1## for a time ##t_0## ##\implies s_{11}= v_1\cdot t_0 ## and then decelerates with ##a## ##\implies s_{12}= \frac{v_1^2}{2a}##. Therefore I assume the condition for no accident is : ##|s_{11}|+ |s_{12}| < d + |s_2|\implies |v_1\cdot t_0|+ |-\frac{1}{2}a(t-t_0)^2| < d + |-\frac{1}{2}at^2|##. After putting in the known values the value of is ##t > 0.362s##. After a little bit of reflection this doesn't make any sense to me - shouldn't be the time bigger?
2. When does the crash happen?
I assumed the distance driven by the car is bigger than the ( distance d + the distance driven by the truck):
##|s_{11}|+ |s_{12}| = d + |s_2|## (so both vehicles are at the same position)
After plugging the values in I get ##t= 0.362s##.

I feel like I missed something here.

Will be grateful for any help.
 
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  • #2
I don't understand this. The truck brakes but then what? Does the car brake ##0.5s## later or not? What time are you trying to calculate in part 1).

What are you assuming for part 2)?
 
  • #3
PeroK said:
I don't understand this. The truck brakes but then what? Does the car brake ##0.5s## later or not? What time are you trying to calculate in part 1).

What are you assuming for part 2)?
I corrected my solving process. The problem does not give any more information so I assumed the truck is stopping until its velocity is 0.
 
  • #4
bolzano95 said:
1. Calculate the time necessary so that the car and the truck don't crash.

What does this mean? Does it mean: how long can the car wait before braking?
 
  • #5
bolzano95 said:
##|s_{11}|+ |s_{12}| < d + |s_2|\implies |v_1\cdot t_0|+ |-\frac{1}{2}a(t-t_0)^2| < d + |-\frac{1}{2}at^2|##.
What is t here? It did not feature in your expression for s12.
 
  • #6
haruspex said:
What is t here? It did not feature in your expression for s12.
t is the particular time in which the truck goes from its velocity ##v_2## to 0 and makes a distance ##s_2##. In the inequality there is difference implemented ##t'= t-t_0##, because in the time t' the car is slowing down.
 
  • #7
PeroK said:
What does this mean? Does it mean: how long can the car wait before braking?
2. The instruction is written here as it is written in my problem book. I personally assume there has to be a condition for the crash not to happen, but I do not know which or what.
 
  • #8
bolzano95 said:
2. The instruction is written here as it is written in my problem book. I personally assume there has to be a condition for the crash not to happen, but I do not know which or what.
A crash seems inevitable to me. The car is too close and driving too fast.
 

1. When will the crash happen?

The exact timing of a crash is difficult to predict, as it depends on many factors such as market conditions, economic indicators, and company performance. However, crashes tend to occur when there is a significant decline in stock prices over a short period of time.

2. What causes a crash to happen?

Crashes can be caused by a variety of factors, including economic recessions, political instability, natural disasters, and unexpected events such as terrorist attacks. They can also be triggered by market speculation, overvaluation of stocks, or excessive borrowing and lending.

3. How can we prevent a crash from happening?

It is impossible to completely prevent a market crash, as they are a natural part of the economic cycle. However, some measures can be taken to mitigate the impact of a crash, such as diversifying investments, setting stop-loss orders, and staying informed about market trends and potential risks.

4. What happens after a crash?

After a crash, there is typically a period of economic downturn and uncertainty. Stock prices may continue to decline, and there may be widespread panic and fear in the market. However, the market eventually stabilizes and begins to recover, although the recovery period can vary in length.

5. Can a crash be predicted?

While it is not possible to predict the exact timing or severity of a crash, there are certain warning signs that can indicate a potential market downturn. These can include high levels of market speculation, excessive borrowing and lending, and overvalued stocks. However, even with these indicators, it is difficult to accurately predict a crash.

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