Question

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?

Answer 1

The student should accelerate and travel through the intersection.

Step-by-step explanation:

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.