Chosing to stop or accelerate through a traffic light

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A driver approaching a yellow traffic light at 47 km/h must decide whether to stop or accelerate. The car's maximum deceleration is -5.6 m/s², and it can accelerate to 70 km/h in 6.9 seconds. Calculations reveal that if the driver brakes, they will travel a certain distance before stopping, while accelerating allows them to cover a distance before the light turns red. The discussion includes solving equations for both scenarios, with participants sharing their calculations and seeking clarification on the correct formulas. Ultimately, the focus is on determining the safest option based on the driver's distance from the intersection and the timing of the traffic light.
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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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