created at May 30, 2021

Can you speed up?

Consider situation: you are driving to the main road, and you see another car is driving. Say, this is the highway, and you can expect that the other car is driving at speed around 100 km/h (see the image below - your car is red). After you are on the road, you start to accelerate immediately, and after a while you will drive at the same speed as the car behind you (100 km/h). Consider also, that there will be a safe distance L between the cars. Also, the possible acceleration of your car is given.

Question:

What should be the minimal initial safe distance L0 between the cars, so that after acceleration up to V2 the distance L will be safe?

Given values are:

V2the other car's speed
V1your car's initial speed
Lsafe distance after acceleration
Ttime for your car to accelerate to 100 km/h

Solving:

Consider your car is moving with a constant acceleration a. If you know the car can hit 100 km/h in T seconds, acceleration a can be calculated as:

a=1003.6T[ms2]a = \frac{100}{3.6\cdot T}\quad \left [ \frac{m}{s^{2}} \right ]

You are starting to accelerate at speed V0 (let's consider it 0), the other car is driving at constant speed V2. After acceleration your speed will be equal to V2.

The time t needed to accelerate is:

t=V2a[s]t = \frac{V_{2}}{a}\quad [s]

The distance L1 your car will travel after acceleration is:

L1=V1t+at22L_{1} = V_{1}\cdot t + \frac{a\cdot t^{2}}{2}

The other car will travel the distance:

L2=V2tL_{2} = V_{2}\cdot t

The image above will help to understand, that:

L0=L2L1+L=(V2V1)tat22+LL_{0} = L_{2} - L_{1} + L = (V_{2} - V_{1})\cdot t - \frac{a\cdot t^{2}}{2} + L

Of coarse, it is the case when you will drive at full throttle. In real life you probably wont push the gas pedal to the floor, but still will sustain quite dynamic acceleration. We could introduce some factor k, which will decrease the maximum possible acceleration a for your car. I would assume it to be about 1.5. I will add it to the Dysolve calculation model.