Safe Distance

Relevant wiki: 1D Kinematics Problem Solving

From the figure above, we can see that

$d_{cb} + d_0 = d_{tb} + d_{tr}$

where

• $d_{cb}$ is the car braking distance $=40 \text{ m}$ .
• $d_0$ , the safety distance we need to calculate.
• $d_{tb}$ is the truck braking distance $=100 \text{ m}$ .
• $d_{tr}$ , the distance traveled by the truck during the reaction time of $\tau = 1.2 \text{ s}$ .

During the reaction time, the truck is travelling at $v_0 = 90 \text{ km/h} = \dfrac {90000}{60 \times 60} = 25 \text{ m/s}$ . Therefore, the distance traveled by the truck during the reaction time is $d_{tr} = v_0 \tau = 25 \times 1.2 = 30 \text{ m}$ . And we have $d_{cb} + d_0 = d_{tb} + d_{tr}$ $\implies d_0 = d_{tb} + d_{tr} - d_{cb} = 100+30-40 = \boxed{90} \text{ m}$ .

90km/h = 90000m per 3600 seconds = 25m/s

So the truck driver would travel:

(1.2 * 25)m = 30 meters in the time taken to react, and would need at least this much extra distance to slow down, so: 100 + 30 = 130 meters.

BUT the car is also moving during the process, 40 meters on it's own. 130 meters only applies if the car comes to an instant dead stop. While the truck still stops after travelling 130m, relative to the car by the end of it's deceleration, it has only travelled:

130 - 40 = 90m

Relevant wiki: 1D Kinematics Problem Solving

Since the car slows down faster and the truck driver reacts with a delay, the truck is always faster than the car, so that the distance between the two vehicles always decreases. Therefore, it is sufficient to consider the final state in which both vehicles have come to a standstill, since in this case the vehicle distance is minimal. During the braking process, the car sets back a distance $s_1$ . Within the reaction time $\tau$ the truck continues to run at a constant speed $v_0$ and, in addition to the car, covers the distance $v_0 \cdot \tau$ in addition to the braking distance $s_2$ . In order in prevent a collision, the safety distance $d_0$ must correspond to the distance the truck additionally covers relative to the car: $d_0 = s_2 - s_1 + v_0 \cdot \tau = 60\,\text{m} + 90 \, \frac{1000\,\text{m}}{3600\,\text{s}} \cdot 1.2 \,\text{s} = 90\,\text{m}$ The other case is more difficult to calculate because the rear vehicle comes to a halt before the first. At the beginning of the braking process, the truck is slower, because the driver of the car only reacts with a delay and decelerates later, so that the vehicle distance is reduced. But as the car slows down much faster, it comes to a halt before the truck, so that the vehicle distance at the end of the braking process increases again. The minimum distance of the vehicle is reached when both have the same speed. We assume that when braking, the speed $v$ decreases linearly with time, so that there is a constant deceleration $a$ . The distance traveled thus depends quadratically on the time: $\begin{aligned} v(t) &= v_0 -a t \\ x(t) &= x_0 + \int_{0}^{t} v(t') dt' = x_0 + v_0 t - \frac{a}{2} t^2 \end{aligned}$ After a time $t = T$ the vehicle has come to a standstill. With the condition $v = 0$ , the braking distance $s$ can be determined: $\begin{aligned} v &= v_0 - a T \stackrel{!}{=} 0 & \Rightarrow \quad T &= \frac{v_0}{a} \\ \Rightarrow \quad s &= x(T) - x(0) = v_0 T - \frac{a}{2} T^2 = \frac{v_0^2}{2 a} & \Rightarrow \quad a &= \frac{v_0^2}{2 s} \\ & & \Rightarrow \quad T &= \frac{v_0}{a} = \frac{2 s}{v_0} \end{aligned}$ This results in braking times of $T_1 = 3.2 \, \text{s}$ for the car and $T_2 = 8 \, \text{s}$ for the truck. Within the time range $t \in [\tau, \tau + T_2]$ , the time laws result $\begin{aligned} x_1(t) &= v_0 t - \frac{v_0^2}{4 s_1} (t - \tau)^2\,, & v_1(t) &= \frac{dx_1}{dt} = v_0 - \frac{v_0^2}{2 s_1} (t - \tau) \\ x_2(t) &= l + d_0 + v_0 t - \frac{v_0^2}{4 s_2} t^2\,, & v_2(t) &= \frac{dx_2}{dt} = v_0 - \frac{v_0^2}{2 s_2} t \end{aligned}$ with the length $l$ of the truck. The condition $v_1 = v_2$ leads to the time $t^\ast$ , where the vehicle distance is minimal: $\begin{aligned} v_0 - \frac{v_0^2}{2 s_1} (t^\ast - \tau) &= v_0 - \frac{v_0^2}{2 s_2} t^\ast \\ \Rightarrow \quad t^\ast &= \frac{s_2}{s_2 - s_1} \tau \end{aligned}$ The minimal vehicle distance results $\begin{aligned} d(t^\ast) &= x_2(t^\ast) - x_1(t^\ast) - l \\ &= d_0 - \frac{v_0^2}{4 s_2} (t^\ast)^2 + \frac{v_0^2}{4 s_1} (t^\ast - \tau)^2 \\ &= d_0 - \frac{s_2}{4 (s_2 - s_1)^2} v_0^2 \tau^2 + \frac{s_1}{4 (s_2 - s_1)^2} v_0^2 \tau^2 \\ &= d_0 - \frac{v_0^2 \tau^2}{4 (s_2 - s_1)} \end{aligned}$ The condition $d (t^\ast) \geq 0$ results in the minimum safety distance to $d_0 = \frac{v_0^2 \tau^2}{4 (s_2 - s_1)} = 3.75 \,\text{m}$ However, you should not rely on the fact, that the truck in front of you decelarates very slowly. You should keep but at least a distance $v_0 \tau = 30 \, \text {m}$ to the front vehicle.

Wow - some folk really make it unnecessarily complicated ! KISS - just do it in your head ... how far does the truck travel in the 1.2 second reaction time ? Answer = 1.2x90000/3600 = 1.2x900/36 = 900/30 = 30 metres. If the truck takes 60 metres more than the car to stop, then the answer is very straightforward : 30 metres to react + 60 metres to brake, so answer - 90 metres.

vc = vt = 90 km/h = 25 m/s

time for car to stop = ts = 40 m / 25 m/s =1.6 s

Total time for truck to stop = tt = 100 m / 25 m/s + 1.2 s = 5.2 s

time taken to travel d0 meter is = tt - tc = 5.2 s - 1.6 s = 3.6 s

d0 = 3.6s * 25 m/s = 90 m

The car will travel 40m. So we know that the distance between car and truck would be 40 m plus the given distance also we have been given the reaction time of truck driver so according to that reaction time we can find the distance that the truck will take until it stops by simple calculation we will get that distance to be 30 M. Now in this particular case the stopping distance for the truck becomes 30 +(stopping distance that is) 100 m. Now we know that to get the minimum distance between two vehicles we must equate 40 m+d°=100+30 In solving this we will get d°=90 m That's the answer.

Look at the car and the truck separately in this problem. The car and the truck travels at the same speed, hence this being possible.

The car would take 40 meters to stop, so the distance between the car and the truck would be (x + 40 meters).

The truck, however, does take 1.2 seconds to react to the car's braking. If you convert 90 km/h to m/s, it would be 25 m/s. So if you take 1.2 * 25, the truck would have to be an additional 30 meters away from the car to eliminate the liability from his reaction time.

The truck takes 100 meters to come to a complete stop and the car takes 40 meters. You can now continue to see the equation as one entity to solve the problem. We know that the difference between the car and truck's stopping distance is 60 meters, so the truck would have to be an additional 60 meters behind the car to start off with. Adding up the reaction distance of 30 meters, the total would then be the only two factors added together, totaling 90 meters.