Is a Crash Inevitable Given the Conditions Described?

[bolzano95]

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.

 

[PeroK]

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)?

 

[bolzano95]

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.

 

[PeroK]

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?

 

[haruspex]

$|s_{11}|+ |s_{12}| < d + |s_2|\implies |v_1\cdot t_0|+ |-\frac{1}{2}a(t-t_0)^2| < d + |-\frac{1}{2}at^2|$.

What is t here? It did not feature in your expression for s₁₂.

 

[bolzano95]

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.

 

[bolzano95]

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.

 

[PeroK]

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.