Solve Retardation Problem: 25m from Traffic Light

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SUMMARY

The discussion centers on solving a physics problem involving a vehicle approaching a traffic light 400 meters away, initially traveling at 15 m/s and decelerating at 0.3 m/s². The correct distance from the traffic light after 1 minute is determined to be 25 meters, based on the vehicle coming to a complete stop after 50 seconds. The confusion arises from misapplying the kinematic equations, which assume continued motion after stopping, rather than modeling the vehicle's deceleration accurately.

PREREQUISITES
  • Understanding of kinematic equations, specifically s = ut + 1/2at²
  • Knowledge of acceleration and deceleration concepts in physics
  • Ability to interpret motion graphs and time intervals
  • Familiarity with the concept of stopping distance in vehicle dynamics
NEXT STEPS
  • Study the application of kinematic equations in real-world scenarios
  • Learn about modeling motion with piecewise functions
  • Explore the concept of stopping distance and its factors in vehicle dynamics
  • Investigate the effects of varying deceleration rates on stopping distance
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding vehicle motion dynamics and braking scenarios.

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Homework Statement


A driver applies brakes on seeing a traffic signal 400 m ahead. At the time of applying brakes the vehicle was moving with 15m/s and retarding with 0.3m/s2. The distance of the vehicle after 1 min from the traffic light is?
A) 25m B) 375m C) 360m D) 40m

Homework Equations

The Attempt at a Solution


Using v=u+at I found the time that the car takes to come v=0 is 50 sec. And using s=ut+1/2at2 I found the distance it traveled in 50 sec is 375m . Hence distance from the traffic light is 25m.
The same problem , this time using s=ut+1/2at2 and t=60 secs (this is what i did first) , i got s= 360m and distance from traffic light =40m. Why exactly is this not working? Is the reason that the equation gives me this answer is that the equation assumes that the car is going to stop at t=50 sec and then continue speeding up in the opposite direction? Please help.
 
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takando12 said:
Is the reason that the equation gives me this answer is that the equation assumes that the car is going to stop at t=50 sec and then continue speeding up in the opposite direction? Please help.
Yes, exactly.

You are expected to model the situation of a person driving up to a stop light. (People don't reverse at stop lights).

Your model was: acceleration = -0.3 m/s/s

But the better model is: For time 0<t<T acceleration = -0.3 m/s/s; for time t>T acceleration = 0 (where T is the time to come to a stop).
 
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Nathanael said:
Yes, exactly.

You are expected to model the situation of a person driving up to a stop light. (People don't reverse at stop lights).

Your model was: acceleration = -0.3 m/s/s

But the better model is: For time 0<t<T acceleration = -0.3 m/s/s; for time t>T acceleration = 0 (where T is the time to come to a stop).

thank you sir,
 

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