Consider a horizontal road with such a set of traffic lights, each spaced 100 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 secondsthe 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 kg and the acceleration of gravity is−9.8 m/s2.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 m/s2.The driver tries to go as fast as possible, while staying at or below the speed limit of 60 km/hr.If the driver sees a red light, they will brake (at 4 m/s2) such that they stop directly under the light.
its simple time = distance/speed
speed = 60km/hr = 60*5/18 m/s
t = 100*3/50
= 6Secs
To find the maximum time t in seconds that the next light must turn green by so the driver of the car does not start braking, we need to consider the distance the car travels during acceleration and deceleration.
Let's break down the problem into different parts:
1. Acceleration phase:
The car accelerates at a rate of 4 m/s^2 until it reaches its maximum speed. We can calculate the time it takes to reach the maximum speed using the following kinematic equation:
v = u + at
Where:
v = final velocity (maximum speed) = 60 km/hr = 16.67 m/s (converted from km/hr to m/s)
u = initial velocity = 0 m/s
a = acceleration = 4 m/s^2
t = time
Rearranging the equation, we have:
t = (v - u) / a
t = (16.67 - 0) / 4
t = 4.17 seconds
So, it takes approximately 4.17 seconds for the car to reach its maximum speed.
2. Distance covered during acceleration:
Using the kinematic equation:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity = 0 m/s
a = acceleration = 4 m/s^2
t = time (4.17 seconds)
We can calculate the distance covered during the acceleration phase:
s = 0*4.17 + (1/2)*4*(4.17)^2
s = 0 + (1/2)*4*(17.32)
s = 34.64 meters
3. Deceleration phase:
Assuming the car's speed remains constant at 60 km/hr during the green light, we need to calculate the distance covered during the deceleration phase until the car comes to a stop.
Using the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s)
u = initial velocity (60 km/hr = 16.67 m/s)
a = deceleration (-4 m/s^2)
s = distance
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
s = (0^2 - (16.67)^2) / (2*-4)
s = (-277.77) / (-8)
s = 34.72 meters
So, the car covers approximately 34.72 meters during the deceleration phase until it comes to a stop.
4. Total distance traveled during a green light cycle:
During a green light, the car covers the distance covered during acceleration (34.64 meters) and the distance covered during deceleration (34.72 meters). Therefore, the total distance traveled during each green light cycle is:
Total distance = Distance during acceleration + Distance during deceleration
Total distance = 34.64 meters + 34.72 meters
Total distance = 69.36 meters
5. Maximum time for the next light to turn green:
To find the maximum time t in seconds that the next light must turn green by so the driver of the car does not start braking, we need to calculate the time it takes for the car to cover the total distance traveled during a green light cycle (69.36 meters), given its constant speed of 60 km/hr.
To convert the speed from km/hr to m/s:
60 km/hr = 60 * (1000/3600) m/s = 16.67 m/s
Using the kinematic equation:
s = ut + (1/2)at^2
Where:
s = distance (69.36 meters)
u = initial velocity (16.67 m/s)
a = acceleration (0 m/s^2 - the car is not accelerating or decelerating during this phase)
t = time for the next light to turn green (maximum time we are trying to find)
Rearranging the equation, we have:
t = (2s) / u
t = (2 * 69.36) / 16.67
t = 8.27 seconds
So, the maximum time t in seconds that the next light must turn green by so the driver of the car does not start braking is approximately 8.27 seconds.