Green waves

Classical Mechanics Level 3

A green wave is a synchronized pattern of traffic lights that helps reduce gridlock. When there is a major thoroughfare with a lot of traffic lights, often the lights turn green sequentially so that cars can travel a long way before they have to stop. This is a "wave of green lights," or green wave.

Consider a horizontal road with such a set of traffic lights, each spaced $100~\mbox{m}$ apart. A car sits at the first traffic light. At $t=0$ the light turns green and the car accelerates. What is the maximum time $t$ in seconds the next light must turn green by so the driver of the car does not start braking?

Details and assumptions

The car has a total mass of $1000~\mbox{kg}$ and the acceleration of gravity is $-9.8~\mbox{m/s}^2$ .
There is no delay between when a light changes color and the driver starts to accelerate/brake (ignore human reaction time).
The car accelerates/brakes at $4~\mbox{m/s}^2$ .
The driver tries to go as fast as possible, while staying at or below the speed limit of $60~\mbox{km/hr}$ .
If the driver sees a red light, they will brake (at $4~\mbox{m/s}^2$ ) such that they stop directly under the light.

The answer is 6.

6 solutions

Devansh Agrawal
May 20, 2014

First, we convert 60 km/h to m/s: $60 km/h = 50/3 m/s$

Then notice that distance taken for the car to accelerate to a maximum speed (of 50/3 m/s) is:

$v^2 = u^2 + 2as$

$(\frac {50}{3} )^2 =0^2 + 2\cdot4 \cdot x$

$x = \frac{625}{18}$

Note that this is the same distance as that of what would be required to slow down. So the distance that is travelled at a constant speed (of 50/3 m/s) is:

$100 - 2\cdot \frac{625}{18} = \frac {275}{9}$

Thus the time taken to travel to the point *before * the car slows down is:

Time to accelerate + time travelling at max speed: Time to accelerate = $50/3 = 4\cdot t$

$t = \frac{25}{6}$

Time travelling at max speed: $x = vt$

$\frac{275}{9} = \frac{50}{3} \cdot t$

$t = \frac{11}{6}$

Therefore the total time is:

$t = \frac{25}{6} + \frac{11}{6} = \frac{36}{6} = 6$

Therefore the answer is 6 seconds.

Jingyi Zhou
May 20, 2014

a = 4m/s^2 v = 60km/hr = 16.67m/s

T (time it takes to accelerate to 16.67 m/s) = (16.67m/s)/(4m/s^2) = 4.167 secs

X (distance traveled during acceleration) = \frac {1}{2}aT^2 = 34.722 m X (distance traveled during braking) = 34.722 m X (distance traveled at 16.67 m/s) = 100m - 34.722m * 2 = 30.556 m

Maximum T so driver doesn't start braking = T (time it takes to accelerate to 16.67 m/s) + T (time it takes to travel 30.556 m at 16.67 m/s) = 4.167 secs + 1.833 = 6 seconds

Kunal Singh
May 20, 2014

The maximum speed limit of the car is 60 km/hr or \frac{50}{3}m/s.So,at any time,the car cannot cross this speed.First,we calculate the time taken to reach this speed.Substituting the values u=0,v=\frac{50}{3}m/s and a=4m/{s^2} in the equation v=u+at,we get t=\frac{25}{6} seconds.Here,we also calculate the distance covered by the car in this time.Substituting the same values of u,v and a in the equation v^2-u^2=2as,we get s=\frac{625}{18}m.Since,the magnitude of acceleration and deceleration(during braking) is the same,the distance travelled by the car during accelerating from zero speed to maximum speed will be the same as during decelerating from maximum speed to zero speed.As the driver has to brake such that he stops under the light after he sees a red light,he has to start braking at a distance of \frac{625}{18}m before the next light.The total distance travelled during acceleration to maximum speed and deceleration to zero speed=2\times\frac{625}{18}m=\frac{625}{9}m,which is not equal to the distance between the lights,i.e. 100m.This means the remaining distance has to be covered by the car under constant maximum speed.The remaining distance=100m-\frac{625}{9}m=\frac{275}{9}m.Time taken to cover this distance=(\frac{275}{9]}m)/(\frac{50}{3}m/s)=\frac{11}{6}seconds.Therefore,the maximum time uptill which the driver does not start braking=(time taken in accelerating from zero speed to maximum speed)+(time taken while travelling at constant maximum speed) =\frac{25}{6} seconds+\frac{11}{6} seconds= 6 seconds.

Bùi Kiên
May 20, 2014

Call the position of the car at first traffic light is A, and the next light is B. We divide the road into 3 segments: AM: the car accelerates. At M (the time is t_1), the car gets maximum speed. *MN: the car moves with a uniform speed v_2 = 60 km/hr = 16.66 m/s. At N (the time is t_2), the drivers will sees the next light turns green. *NB: the car moves with a uniform speed v_3 = v_2. According the last assumption, it's easy to prove that AM = NB. We have t_1 = v_{max}/a = 16.67/4 = 4.1675 s and AM = a t 1 ^2 /2 = 34.7 m. Hence MN = AB - 2*AM = 30.6(m) and t 2 = 1.836 s. The result is t = t 1 + t 2 = 6.001 s.

Víctor Martín
Jan 6, 2014

The car can move at 60km/h. Dividing by 3.6 we get 16.666 m/s. The distance between traffic lights is 100 m, so $\frac{100}{16.666} = \boxed{6}$

No! I think your solution is wrong even though the answer is correct. You should consider that the car starts at rest. If the car moves, it can only be accelerated until the velocity is $60 \,\text{km/h}$ due to the speed limit. Therefore, the distance $s$ that the car covers and the time that is spent to reach speed limit $60 \,\text{km/h}$ or $\frac{50}{3} m/s$ are $s=\frac{v_t^2-v_o^2}{2a}=\frac{\left(\frac{50}{3}\right)^2-0^2}{2\cdot 4}=\frac{625}{18}m$ and $t_1=\frac{v_t-v_o}{a}=\frac{\frac{50}{3}-0}{4}=\frac{25}{6}s.$ After this state, the the car will move at constant speed $60 \,\text{km/h}$ . The condition that must be required to avoid the driver hitting the brakes is the traffic light must turn green when the distance between the car and the traffic light is $\frac{625}{18}m$ . Hence, the distance that car covers at constant speed $60 \,\text{km/h}$ is $100-2\cdot \left(\frac{625}{18}\right)= \frac{275}{9}m$ and the time that is spent to cover that distance is $t_2=\frac{\frac{275}{9}}{\frac{50}{3}}=\frac{11}{6}s.$ Thus, the maximum time is $t_1 + t_2 = \frac{25}{6} + \frac{11}{6} = \boxed{6\,s}$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

Tunk-Fey Ariawan - 7 years, 4 months ago

Enrique Vizcarra
Dec 21, 2013

If he wants to respect the velocity limit then the maximum velocity (Vmax) he will reach will be 60km/hr, that is 60/3.6 m/s= 50/3 m/s >>Vmax. So the time it takes to get to that speed starting from repose can be find using Vf=Vi+at, so for this case : 50/3= 0 + 4t this gives a time of 25/6 s>> t1 Well after that time it will be at position (using Xf-Xi=vit + 1/2at^2): X= 1/2a t^2= 1/2 (4)(25/6)^2= 625/18 m>>x. we should keep in mind the same time it takes to get to that speed is the same time it takes to completely stop starting from Vmax if using same acceleration's magnitude to brake, and analogously it will advance the same distance of x while braking, so that let us 100m - 2x distance to move at Vmax, that is 100m-625/9 m =275/9m >>d. So those 275/9m are the middle distance it moves at constant Vmax= 50/3 m/s (because we're respecting the limit we are not accelerating no more) then that takes a time of (using t= d/v >>t2 = d/Vmax) t2= (275/9 m)/(50/3 m/s)= 11/6 s. So that total time it pass to reach the point where it is suppose to start braking is t1 +t2 = 25/6s + 11/6s = 36/6s= 6s

Green waves

The answer is 6.

6 solutions

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