Chosing to stop or accelerate through a traffic light

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Homework Help Overview

The discussion revolves around a scenario involving a driver approaching a traffic light that has just turned yellow. The driver must decide whether to stop or accelerate through the intersection, considering the distance to the light and the time it will remain yellow. The problem involves concepts from kinematics, including acceleration, deceleration, and distance traveled under varying conditions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations of motion to determine stopping distance and the distance traveled if accelerating. There are attempts to apply the kinematic equation V^2 = Vi^2 + 2ax, with some participants questioning the correctness of their calculations and assumptions. Others explore the implications of the driver's choices and the time constraints imposed by the traffic light.

Discussion Status

The discussion includes various attempts to solve the problem, with some participants expressing confusion over their calculations. There is acknowledgment of the need to clarify the application of equations, and some guidance has been offered regarding the correct setup for the calculations. The conversation is ongoing, with multiple interpretations being explored.

Contextual Notes

Participants note the constraints of the problem, including the specific distances and times provided, as well as the maximum deceleration and acceleration of the vehicle. There is also a separate question introduced regarding a rock dropped from a cliff, indicating a shift in focus to another physics problem.

sadeysnow
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1. A person driving her car at 47 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 30 m away from the near side of the intersection (Fig. 2-29). Should she try to stop, or should she make a run for it? The intersection is 15 m wide. Her car's maximum deceleration is -5.6 m/s2, whereas it can accelerate from 47 km/h to 70 km/h in 6.9 s. Ignore the length of her car and her reaction time.

If she hits the brakes, how far will she travel before stopping?
If she hits the gas instead, how far will she travel before the light turns red?


2. V^2=Vi^2+2ax



3. 47 km/h=13.05 m/s 0=(13.05)^2+29-5.6)=159.1 (That just doesn't seem right what am I doing wrong)
 
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Please rewrite the equation. " (13.05)^2+29-5.6) " is not correct.

Can the driver stop in 30 m or less?

V^2=Vi^2+2ax is correct. Solve for x knowing that v = 0, vi = 13.05 m/s and a = -5.6 m/s.

I get an answer an order of magnitude less than 159.1.
 
I do that and it doesn't work its the wrong answer everytime...
 
x=v^2/(2a)

v=13.05
a=5.6

and what do u get for x now?
 
thanks...I got the answer I needed.
 
what about the second part of the problem?
 
I guess you haven't showed any work for it =XP
 
ok what I did was:

Vi=13.05 Vf=26.38 t=6.9 a=?
vf=vi+at

26.38=13.05+(a)(6.9) = 1.93

then I used.
d=vi(t)+.5at^2

d=13.05(2)+.5(1.93)(2)^2=29.96

and that's not the right answer...
 
sadeysnow said:
ok what I did was:

Vi=13.05 Vf=26.38 t=6.9 a=?

70 km/h = 19.4444444 meters / second
 
  • #10
thanks!
 
  • #11
np
:smile:
 
  • #12
A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 4.2 s later. If the speed of sound is 340 m/s, how high is the cliff?


I don't even know where to start...can you help?
 
  • #13
the stone striked the ground after 't' seconds,
and, then the sound starts traveling at v of 340 m/s (I guess without the influence of gravity), and reaches to the top in " t' " seconds

and, you know t+t'=4.2 s,
 

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