A physics student is driving home after class. The car is traveling at 14.7 m/s when it approaches an intersection. The student estimates that he is 20.0 m from the entrance to the intersection when the traffic light changes from green to yellow and the intersection is 10.0 m wide. The light will change from yellow to red in 3.00 s. The maximum safe deceleration of the car is 4.00 m/s2 while the maximum acceleration of the car is 2.00 m/s2. Should the physics student decelerate and stop or accelerate and travel through the intersection?

As per the given problem, the velocity with which the car of the student is moving is 14.7 m/s and he is 20 m away from intersection. Also the width of intersection is 10 m. It is also stated that traffic lights will change from yellow to red in 3 s. So totally , the student has to cover a distance of 30 m in 3 s to avoid getting stopped in intersection.

The velocity or speed required by the car to cover 30 m in 3 s is 30/3 = 15 m/s.

And the car is moving with initial velocity of 14.7 m/s. The student has two options either to decelerate at 4m/s² rate or to accelerate at the rate of 2 m/s².

So the velocity or speed of the car in 3 s with acceleration of 2 m/s² will be

Velocity = Acceleration × Time = 2 × 3 =6 m/s.

Thus on accelerating the car by 2 m/s² in 3 s, it will have the speed of 6 m/s and it can cover a distance of 6 × 3 = 18 m.

And the time taken by the car to reach from 20 m to intersection point with speed of 14.7 m/s is 20/14.7 = 1.36 s.

So, the car will take 1.36 s to reach the entrance of intersection with the speed of 14.7 m/s. But if it gets accelerated to 2 m/s², then it will have an increase in speed which will make the car to cover the complete intersection without stopping. So the student should accelerate the car.